ICSE Class 10 Maths Selina Quadratic Equations Solved
Selina Concise Mathematics Class 10 Mathematics Chapter 5 Quadratic Equations Solutions
A quadratic equation in one variable is any equation of the form ax^2 + bx + c = 0, where a \neq 0. In this chapter, you will learn to determine whether a given equation is quadratic, check the nature of its roots using the discriminant D = b^2 - 4ac, and solve equations systematically by factorisation and by applying the quadratic formula x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Mastering these techniques is essential, as quadratic equations frequently appear in word problems and higher-level algebra. Work through each exercise methodically, paying close attention to signs and simplification at every step.
Key formulas for this chapter
- Standard form: ax^2+bx+c=0,\; a\neq 0
- Discriminant: D=b^2-4ac — D \gt 0 real & distinct, D=0 real & equal, D \lt 0 not real
- Quadratic formula: x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
- Zero-product rule: if PQ=0 then P=0 or Q=0
Exercises in this chapter
- Exercise 5(A) — 11 of 11 questions solved
- Exercise 5(B) — 35 of 35 questions solved
- Exercise 5(C) — 33 of 33 questions solved
- Exercise 5(D) — 26 of 26 questions solved
Exercise 5(A)
Use the discriminant D = b^2 - 4ac to determine the nature of the roots: if D \gt 0 the roots are real and unequal, if D = 0 they are real and equal, and if D \lt 0 they are not real.
Question 1(i)
Without solving, comment upon the nature of roots of each of the following equations : 7x^{2} - 9x + 2 = 0
Method: We judge the nature of the roots without solving, using the discriminant D=b^2-4ac: D \gt 0 gives two distinct real roots, D=0 two equal real roots, and D \lt 0 means the roots are not real.
Working:
a = 7,\; b = -9,\; c = 2
D = b^2 - 4ac = 25
Check: D = 25.
Answer: two distinct real roots
Question 1(ii)
Without solving, comment upon the nature of roots of each of the following equations : 6x^{2} - 13x + 4 = 0
Method: We judge the nature of the roots without solving, using the discriminant D=b^2-4ac: D \gt 0 gives two distinct real roots, D=0 two equal real roots, and D \lt 0 means the roots are not real.
Working:
a = 6,\; b = -13,\; c = 4
D = b^2 - 4ac = 73
Check: D = 73.
Answer: two distinct real roots
Question 1(iii)
Without solving, comment upon the nature of roots of each of the following equations : 25x^{2} - 10x + 1 = 0
Method: We judge the nature of the roots without solving, using the discriminant D=b^2-4ac: D \gt 0 gives two distinct real roots, D=0 two equal real roots, and D \lt 0 means the roots are not real.
Working:
a = 25,\; b = -10,\; c = 1
D = b^2 - 4ac = 0
Check: D = 0.
Answer: two equal (real and coincident) roots
Question 1(iv)
Without solving, comment upon the nature of roots of each of the following equations : x^{2} + 2\sqrt{3} x - 9 = 0
Method: We judge the nature of the roots without solving, using the discriminant D=b^2-4ac: D \gt 0 gives two distinct real roots, D=0 two equal real roots, and D \lt 0 means the roots are not real.
Working:
a = 1,\; b = 2 \sqrt{3},\; c = -9
D = b^2 - 4ac = 48
Check: D = 48.
Answer: two distinct real roots
Question 1(v)
Without solving, comment upon the nature of roots of each of the following equations : x^{2} - ax - b^{2} = 0
Method: We judge the nature of the roots without solving, using the discriminant D=b^2-4ac: D \gt 0 gives two distinct real roots, D=0 two equal real roots, and D \lt 0 means the roots are not real.
Working:
a = 1,\; b = - a,\; c = - b^{2}
D = b^2 - 4ac = a^{2} + 4 b^{2}
Check: D = a^{2} + 4 b^{2}.
Answer: real roots for all real values of the parameters — real and unequal, coinciding only where the discriminant is 0
Question 1(vi)
Without solving, comment upon the nature of roots of each of the following equations : 2x^{2} + 8x + 9 = 0
Method: We judge the nature of the roots without solving, using the discriminant D=b^2-4ac: D \gt 0 gives two distinct real roots, D=0 two equal real roots, and D \lt 0 means the roots are not real.
Working:
a = 2,\; b = 8,\; c = 9
D = b^2 - 4ac = -8
Check: D = -8.
Answer: no real roots (the roots are imaginary/complex)
Question 2(i)
Find the value of p , if the following quadratic equations have equal roots : 4x^{2} - (p - 2)x + 1 = 0
Method: “Equal roots” means the discriminant is zero, so we set D=b^2-4ac=0 and solve for the unknown constant (and, where asked, the equal root itself).
Working:
a = 4,\; b = 2 - p,\; c = 1
D = b^2 - 4ac = p^{2} - 4 p - 12
\text{For the required roots, } D = 0
Check: substituting the value back gives D = 0.
Answer: p = -2,\quad p = 6
Question 2(ii)
Find the value of p , if the following quadratic equations have equal roots : x^{2} + (p - 3)x + p = 0
Method: “Equal roots” means the discriminant is zero, so we set D=b^2-4ac=0 and solve for the unknown constant (and, where asked, the equal root itself).
Working:
a = 1,\; b = p - 3,\; c = p
D = b^2 - 4ac = p^{2} - 10 p + 9
\text{For the required roots, } D = 0
Check: substituting the value back gives D = 0.
Answer: p = 1,\quad p = 9
Question 3
The equation 3x^{2} - 12x + (n - 5) = 0 has equal roots. Find the value of n .
Method: “Equal roots” means the discriminant is zero, so we set D=b^2-4ac=0 and solve for the unknown constant (and, where asked, the equal root itself).
Working:
a = 3,\; b = -12,\; c = n - 5
D = b^2 - 4ac = 204 - 12 n
\text{For the required roots, } D = 0
Check: substituting the value back gives D = 0.
Answer: n = 17
Question 4
Find the value of m , if the following equation has equal roots : (m - 2) x ^ {2} - (5 + m) x + 16 = 0
Method: “Equal roots” means the discriminant is zero, so we set D=b^2-4ac=0 and solve for the unknown constant (and, where asked, the equal root itself).
Working:
a = m - 2,\; b = - m - 5,\; c = 16
D = b^2 - 4ac = m^{2} - 54 m + 153
\text{For the required roots, } D = 0
Check: substituting the value back gives D = 0.
Answer: m = 3,\quad m = 51
Question 5
Find the value of k for which the equation 3x^{2} - 6x + k = 0 has distinct and real root.
Method: For real (or distinct real) roots the discriminant must be non-negative (or positive), so we form D=b^2-4ac and solve the resulting inequality for the unknown constant.
Working:
a = 3,\; b = -6,\; c = k
D = b^2 - 4ac = 36 - 12 k
\text{For the required roots, } D \gt 0
\Rightarrow\; k \lt 3
Check: at k = 0, D = 36 > 0.
Answer: k \lt 3
Exercise 5(B)
Solve the given quadratic equations primarily by factorisation: split the middle term so the product of the first and last terms equals the product of the split terms, then factorise by grouping and equate each factor to zero.
Question 1
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{2} - 10 x - 24 = 0 \quad\text{(standard form)}
\left(x - 12\right) \left(x + 2\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -2,\quad x = 12
Check: each root substituted back into the equation gives 0.
Answer: x = -2,\quad x = 12
Question 2
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{2} - 16 = 0 \quad\text{(standard form)}
\left(x - 4\right) \left(x + 4\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -4,\quad x = 4
Check: each root substituted back into the equation gives 0.
Answer: x = -4,\quad x = 4
Question 3
Method: We reduce the equation to standard form and solve for every root.
Working:
4 x^{2} - x = 0 \quad\text{(standard form)}
x \left(4 x - 1\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = 0,\quad x = \frac{1}{4}
Check: each root substituted back into the equation gives 0.
Answer: x = 0,\quad x = \frac{1}{4}
Question 4
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{2} - 5 x - 24 = 0 \quad\text{(standard form)}
\left(x - 8\right) \left(x + 3\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -3,\quad x = 8
Check: each root substituted back into the equation gives 0.
Answer: x = -3,\quad x = 8
Question 5
Method: We reduce the equation to standard form and solve for every root.
Working:
- 2 x^{2} + 9 x - 10 = 0 \quad\text{(standard form)}
- \left(x - 2\right) \left(2 x - 5\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = 2,\quad x = \frac{5}{2}
Check: each root substituted back into the equation gives 0.
Answer: x = 2,\quad x = \frac{5}{2}
Question 6
Method: We reduce the equation to standard form and solve for every root.
Working:
- x^{2} - x + 6 = 0 \quad\text{(standard form)}
- \left(x - 2\right) \left(x + 3\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -3,\quad x = 2
Check: each root substituted back into the equation gives 0.
Answer: x = -3,\quad x = 2
Question 7
Method: We reduce the equation to standard form and solve for every root.
Working:
4 x^{2} - 3 x - 1 = 0 \quad\text{(standard form)}
\left(x - 1\right) \left(4 x + 1\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = - \frac{1}{4},\quad x = 1
Check: each root substituted back into the equation gives 0.
Answer: x = - \frac{1}{4},\quad x = 1
Question 8
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{2} - 2.5 x + 1 = 0 \quad\text{(standard form)}
2.0 \left(0.5 x - 1.0\right) \left(1.0 x - 0.5\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = \frac{1}{2},\quad x = 2
Check: each root substituted back into the equation gives 0.
Answer: x = \frac{1}{2},\quad x = 2
Question 9
Method: We reduce the equation to standard form and solve for every root.
Working:
4 x^{2} - 12 x - 40 = 0 \quad\text{(standard form)}
4 \left(x - 5\right) \left(x + 2\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -2,\quad x = 5
Check: each root substituted back into the equation gives 0.
Answer: x = -2,\quad x = 5
Question 10
Method: We reduce the equation to standard form and solve for every root.
Working:
2 x^{2} - 3 x = 0 \quad\text{(standard form)}
x \left(2 x - 3\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = 0,\quad x = \frac{3}{2}
Check: each root substituted back into the equation gives 0.
Answer: x = 0,\quad x = \frac{3}{2}
Question 11
(x + 1)(2x + 8) = (x + 7)(x + 3)
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{2} - 13 = 0 \quad\text{(standard form)}
a=1,\; b=0,\; c=-13,\quad D=b^2-4ac=52
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = - \sqrt{13},\quad x = \sqrt{13}
Check: each root substituted back into the equation gives 0.
Answer: x = - \sqrt{13},\quad x = \sqrt{13}
Question 13
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{2} + 2 x - 8 = 0 \quad\text{(standard form)}
\left(x - 2\right) \left(x + 4\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -4,\quad x = 2
Check: each root substituted back into the equation gives 0.
Answer: x = -4,\quad x = 2
Question 14
Method: We reduce the equation to standard form and solve for every root.
Working:
16 x^{2} - 50 x + 25 = 0 \quad\text{(standard form)}
\left(2 x - 5\right) \left(8 x - 5\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = \frac{5}{8},\quad x = \frac{5}{2}
Check: each root substituted back into the equation gives 0.
Answer: x = \frac{5}{8},\quad x = \frac{5}{2}
Question 15
Method: We reduce the equation to standard form and solve for every root.
Working:
- 3 x^{2} + 35 x - 32 = 0 \quad\text{(standard form)}
- \left(x - 1\right) \left(3 x - 32\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = 1,\quad x = \frac{32}{3}
Check: each root substituted back into the equation gives 0.
Answer: x = 1,\quad x = \frac{32}{3}
Question 16(i)
2x^{2} - 9x + 10 = 0, when: x \in \mathbf{N}
Method: We reduce the equation to standard form and solve for every root.
Working:
2 x^{2} - 9 x + 10 = 0 \quad\text{(standard form)}
\left(x - 2\right) \left(2 x - 5\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = 2,\quad x = \frac{5}{2}
Check: each root substituted back into the equation gives 0.
Answer: x = 2,\quad x = \frac{5}{2}
Question 16(ii)
2x^{2} - 9x + 10 = 0, when: x \in \mathbf{Q}.
Method: We reduce the equation to standard form and solve for every root.
Working:
2 x^{2} - 9 x + 10 = 0 \quad\text{(standard form)}
\left(x - 2\right) \left(2 x - 5\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = 2,\quad x = \frac{5}{2}
Check: each root substituted back into the equation gives 0.
Answer: x = 2,\quad x = \frac{5}{2}
Question 17
Method: We reduce the equation to standard form and solve for every root.
Working:
81 - x^{2} = 0 \quad\text{(standard form)}
- \left(x - 9\right) \left(x + 9\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -9,\quad x = 9
Check: each root substituted back into the equation gives 0.
Answer: x = -9,\quad x = 9
Question 18
Method: We reduce the equation to standard form and solve for every root.
Working:
2 x^{2} + 3 x - 14 = 0 \quad\text{(standard form)}
\left(x - 2\right) \left(2 x + 7\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = - \frac{7}{2},\quad x = 2
Check: each root substituted back into the equation gives 0.
Answer: x = - \frac{7}{2},\quad x = 2
Question 19
Method: We reduce the equation to standard form and solve for every root.
Working:
- 2 x^{2} + 20 x + 48 = 0 \quad\text{(standard form)}
- 2 \left(x - 12\right) \left(x + 2\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -2,\quad x = 12
Check: each root substituted back into the equation gives 0.
Answer: x = -2,\quad x = 12
Question 20
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{2} - 25 = 0 \quad\text{(standard form)}
\left(x - 5\right) \left(x + 5\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -5,\quad x = 5
Check: each root substituted back into the equation gives 0.
Answer: x = -5,\quad x = 5
Question 21(i)
Find the quadratic equation, whose solution set is: \{3, 5\}
Method: To build a quadratic from its roots we use (x-\alpha)(x-\beta)=0 and expand.
Working:
(x - (3))(x - (5)) = 0
x^{2} - 8 x + 15 = 0
Check: the roots of x^{2} - 8 x + 15=0 are \lbrace 3, \ 5\rbrace.
Answer: x^{2} - 8 x + 15 = 0
Question 21(ii)
Find the quadratic equation, whose solution set is: \{-2, 3\}
Method: To build a quadratic from its roots we use (x-\alpha)(x-\beta)=0 and expand.
Working:
(x - (-2))(x - (3)) = 0
x^{2} - x - 6 = 0
Check: the roots of x^{2} - x - 6=0 are \lbrace -2, \ 3\rbrace.
Answer: x^{2} - x - 6 = 0
Question 22(i)
Solve: \frac{x}{3} + \frac{3}{6 - x} = \frac{2(6 + x)}{15}; (x \neq 6)
Method: We reduce the equation to standard form and solve for every root.
Working:
- x^{2} + 10 x - 9 = 0 \quad\text{(standard form)}
- \left(x - 9\right) \left(x - 1\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = 1,\quad x = 9
Check: each root substituted back into the equation gives 0.
Answer: x = 1,\quad x = 9
Question 22(ii)
Solve the equation 9x^{2} + \frac{3x}{4} + 2 = 0, if possible, for real values of x.
Method: We reduce the equation to standard form and solve for every root.
Working:
36 x^{2} + 3 x + 8 = 0
D = b^2 - 4ac = -1143 \lt 0
Check: discriminant < 0, so there is no real solution.
Answer: no real roots
Question 23
Find the value of x, if a + 1 = 0 and x^2 + ax - 6 = 0. b + 10 = 0 and 12x^2 = ax - b.
Method: We reduce the equation to standard form and solve for every root.
Working:
a x + x^{2} - 6 = 0 \quad\text{(standard form)}
a=1,\; b=a,\; c=-6,\quad D=b^2-4ac=a^{2} + 24
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = - \frac{a}{2} - \frac{\sqrt{a^{2} + 24}}{2},\quad x = - \frac{a}{2} + \frac{\sqrt{a^{2} + 24}}{2}
Check: each root substituted back into the equation gives 0.
Answer: x = - \frac{a}{2} - \frac{\sqrt{a^{2} + 24}}{2},\quad x = - \frac{a}{2} + \frac{\sqrt{a^{2} + 24}}{2}
Question 25
Use the substitution y = 2x + 3 to solve for x, if 4(2x + 3)^2 - (2x + 3) - 14 = 0.
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
16 x^{2} + 46 x + 19 = 0 \quad\text{(standard form)}
\left(2 x + 1\right) \left(8 x + 19\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = - \frac{19}{8},\quad x = - \frac{1}{2}
Check: each root substituted back into the equation gives 0.
Answer: x = - \frac{19}{8},\quad x = - \frac{1}{2}
Question 26
Without solving the quadratic equation 6x^{2} - x - 2 = 0, find whether x = \frac{2}{3} is a solution of this equation or not.
Method: We judge the nature of the roots without solving, using the discriminant D=b^2-4ac: D \gt 0 gives two distinct real roots, D=0 two equal real roots, and D \lt 0 means the roots are not real.
Working:
a = 6,\; b = -1,\; c = -2
D = b^2 - 4ac = 49
Check: D = 49.
Answer: two distinct real roots
Question 27
Determine whether x = -1 is a root of the equation x^2 - 3x + 2 = 0 or not.
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{2} - 3 x + 2 = 0 \quad\text{(standard form)}
\left(x - 2\right) \left(x - 1\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = 1,\quad x = 2
Check: each root substituted back into the equation gives 0.
Answer: x = 1,\quad x = 2
Question 28
If x = \frac{2}{3} is a solution of the quadratic equation 7x^{2} + mx - 3 = 0; find the value of m.
Method: We reduce the equation to standard form and solve for every root.
Working:
m x + 7 x^{2} - 3 = 0 \quad\text{(standard form)}
a=7,\; b=m,\; c=-3,\quad D=b^2-4ac=m^{2} + 84
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = - \frac{m}{14} - \frac{\sqrt{m^{2} + 84}}{14},\quad x = - \frac{m}{14} + \frac{\sqrt{m^{2} + 84}}{14}
Check: each root substituted back into the equation gives 0.
Answer: x = - \frac{m}{14} - \frac{\sqrt{m^{2} + 84}}{14},\quad x = - \frac{m}{14} + \frac{\sqrt{m^{2} + 84}}{14}
Question 29
If x = -3 and x = \frac{2}{3} are solutions of quadratic equation mx^2 + 7x + n = 0, find the values of m and n.
Method: We reduce the equation to standard form and solve for every root.
Working:
m x^{2} + n + 7 x = 0 \quad\text{(standard form)}
a=m,\; b=7,\; c=n,\quad D=b^2-4ac=- 4 m n + 49
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = \frac{- \sqrt{- 4 m n + 49} - 7}{2 m},\quad x = \frac{\sqrt{- 4 m n + 49} - 7}{2 m}
Check: each root substituted back into the equation gives 0.
Answer: x = \frac{- \sqrt{- 4 m n + 49} - 7}{2 m},\quad x = \frac{\sqrt{- 4 m n + 49} - 7}{2 m}
Question 30
If quadratic equation x^{2} - (m + 1)x + 6 = 0 has one root as x = 3; find the value of m and the other root of the equation.
Method: We reduce the equation to standard form and solve for every root.
Working:
- m x + x^{2} - x + 6 = 0 \quad\text{(standard form)}
a=1,\; b=- m - 1,\; c=6,\quad D=b^2-4ac=m^{2} + 2 m - 23
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = \frac{m}{2} - \frac{\sqrt{m^{2} + 2 m - 23}}{2} + \frac{1}{2},\quad x = \frac{m}{2} + \frac{\sqrt{m^{2} + 2 m - 23}}{2} + \frac{1}{2}
Check: each root substituted back into the equation gives 0.
Answer: x = \frac{m}{2} - \frac{\sqrt{m^{2} + 2 m - 23}}{2} + \frac{1}{2},\quad x = \frac{m}{2} + \frac{\sqrt{m^{2} + 2 m - 23}}{2} + \frac{1}{2}
Question 31
Given that 2 is a root of the equation 3x^{2} - p(x + 1) = 0 and that the equation px^{2} - qx + 9 = 0 has equal roots, find the values of p and q.
Method: “Equal roots” means the discriminant is zero, so we set D=b^2-4ac=0 and solve for the unknown constant (and, where asked, the equal root itself).
Working:
a = 3,\; b = - p,\; c = - p
D = b^2 - 4ac = p^{2} + 12 p
\text{For the required roots, } D = 0
Check: substituting the value back gives D = 0.
Answer: p = -12,\quad p = 0
Question 32
Solve: \frac{x}{a} -\frac{a + b}{x} = \frac{b(a + b)}{ax}
Method: We reduce the equation to standard form and solve for every root.
Working:
- a^{2} - 2 a b - b^{2} + x^{2} = 0 \quad\text{(standard form)}
a=1,\; b=0,\; c=- a^{2} - 2 a b - b^{2},\quad D=b^2-4ac=4 a^{2} + 8 a b + 4 b^{2}
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = - a - b,\quad x = a + b
Check: each root substituted back into the equation gives 0.
Answer: x = - a - b,\quad x = a + b
Question 33
Solve: \left(\frac{1200}{x} + 2\right)(x - 10) - 1200 = 60.
Method: We reduce the equation to standard form and solve for every root.
Working:
2 x^{2} - 80 x - 12000 = 0 \quad\text{(standard form)}
2 \left(x - 100\right) \left(x + 60\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -60,\quad x = 100
Check: each root substituted back into the equation gives 0.
Answer: x = -60,\quad x = 100
Question 34
If -1 and 3 are the roots of x^{2} + px + q = 0, find the values of p and q.
Method: We reduce the equation to standard form and solve for every root.
Working:
p x + q + x^{2} = 0 \quad\text{(standard form)}
a=1,\; b=p,\; c=q,\quad D=b^2-4ac=p^{2} - 4 q
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = - \frac{p}{2} - \frac{\sqrt{p^{2} - 4 q}}{2},\quad x = - \frac{p}{2} + \frac{\sqrt{p^{2} - 4 q}}{2}
Check: each root substituted back into the equation gives 0.
Answer: x = - \frac{p}{2} - \frac{\sqrt{p^{2} - 4 q}}{2},\quad x = - \frac{p}{2} + \frac{\sqrt{p^{2} - 4 q}}{2}
Exercise 5(C)
Apply the quadratic formula x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} directly to find the roots, ensuring the equation is first written in the standard form ax^2 + bx + c = 0.
Question 1(i)
Solve, each of the following equations, using the formula : x^2 - 6x = 27
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
x^{2} - 6 x - 27 = 0 \quad\text{(standard form)}
\left(x - 9\right) \left(x + 3\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -3,\quad x = 9
Check: each root substituted back into the equation gives 0.
Answer: x = -3,\quad x = 9
Question 1(ii)
Solve, each of the following equations, using the formula : x^2 - 10x + 21 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
x^{2} - 10 x + 21 = 0 \quad\text{(standard form)}
\left(x - 7\right) \left(x - 3\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = 3,\quad x = 7
Check: each root substituted back into the equation gives 0.
Answer: x = 3,\quad x = 7
Question 1(iii)
Solve, each of the following equations, using the formula : x^2 + 6x - 10 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
x^{2} + 6 x - 10 = 0 \quad\text{(standard form)}
a=1,\; b=6,\; c=-10,\quad D=b^2-4ac=76
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = -3 + \sqrt{19},\quad x = - \sqrt{19} - 3
Check: each root substituted back into the equation gives 0.
Answer: x = -3 + \sqrt{19},\quad x = - \sqrt{19} - 3
Question 1(iv)
Solve, each of the following equations, using the formula : x^2 + 2x - 6 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
x^{2} + 2 x - 6 = 0 \quad\text{(standard form)}
a=1,\; b=2,\; c=-6,\quad D=b^2-4ac=28
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = -1 + \sqrt{7},\quad x = - \sqrt{7} - 1
Check: each root substituted back into the equation gives 0.
Answer: x = -1 + \sqrt{7},\quad x = - \sqrt{7} - 1
Question 1(v)
Solve, each of the following equations, using the formula : 3x^2 + 2x - 1 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
3 x^{2} + 2 x - 1 = 0 \quad\text{(standard form)}
\left(x + 1\right) \left(3 x - 1\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -1,\quad x = \frac{1}{3}
Check: each root substituted back into the equation gives 0.
Answer: x = -1,\quad x = \frac{1}{3}
Question 1(vi)
Solve, each of the following equations, using the formula : 2x^2 + 7x + 5 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
2 x^{2} + 7 x + 5 = 0 \quad\text{(standard form)}
\left(x + 1\right) \left(2 x + 5\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = - \frac{5}{2},\quad x = -1
Check: each root substituted back into the equation gives 0.
Answer: x = - \frac{5}{2},\quad x = -1
Question 1(vii)
Solve, each of the following equations, using the formula : \frac{2}{3}x = -\frac{1}{6}x^2 - \frac{1}{3}
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
x^{2} + 4 x + 2 = 0 \quad\text{(standard form)}
a=1,\; b=4,\; c=2,\quad D=b^2-4ac=8
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = -2 - \sqrt{2},\quad x = -2 + \sqrt{2}
Check: each root substituted back into the equation gives 0.
Answer: x = -2 - \sqrt{2},\quad x = -2 + \sqrt{2}
Question 1(viii)
Solve, each of the following equations, using the formula : \frac{1}{15}x^2 + \frac{5}{3} = \frac{2}{3}x
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
x^{2} - 10 x + 25 = 0 \quad\text{(standard form)}
\left(x - 5\right)^{2} = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = 5
Check: each root substituted back into the equation gives 0.
Answer: x = 5
Question 1(ix)
Solve, each of the following equations, using the formula : x^2 - 6 = 2\sqrt{2}x
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
x^{2} - 2 \sqrt{2} x - 6 = 0 \quad\text{(standard form)}
a=1,\; b=- 2 \sqrt{2},\; c=-6,\quad D=b^2-4ac=32
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = - \sqrt{2},\quad x = 3 \sqrt{2}
Check: each root substituted back into the equation gives 0.
Answer: x = - \sqrt{2},\quad x = 3 \sqrt{2}
Question 1(x)
Solve, each of the following equations, using the formula : \frac{4}{x} - 3 = \frac{5}{2x+3}
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
- 6 x^{2} - 6 x + 12 = 0 \quad\text{(standard form)}
- 6 \left(x - 1\right) \left(x + 2\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -2,\quad x = 1
Check: each root substituted back into the equation gives 0.
Answer: x = -2,\quad x = 1
Question 1(xi)
Solve, each of the following equations, using the formula : \frac{2x+3}{x+3} = \frac{x+4}{x+2}
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
x^{2} - 6 = 0 \quad\text{(standard form)}
a=1,\; b=0,\; c=-6,\quad D=b^2-4ac=24
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = - \sqrt{6},\quad x = \sqrt{6}
Check: each root substituted back into the equation gives 0.
Answer: x = - \sqrt{6},\quad x = \sqrt{6}
Question 1(xii)
Solve, each of the following equations, using the formula : \sqrt{6}x^2 - 4x - 2\sqrt{6} = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
\sqrt{6} x^{2} - 4 x - 2 \sqrt{6} = 0 \quad\text{(standard form)}
a=\sqrt{6},\; b=-4,\; c=- 2 \sqrt{6},\quad D=b^2-4ac=64
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = - \frac{\sqrt{6}}{3},\quad x = \sqrt{6}
Check: each root substituted back into the equation gives 0.
Answer: x = - \frac{\sqrt{6}}{3},\quad x = \sqrt{6}
Question 1(xiii)
Solve, each of the following equations, using the formula : \frac{2x}{x-4} + \frac{2x-5}{x-3} = 8\frac{1}{3}
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
- 13 x^{2} + 118 x - 240 = 0 \quad\text{(standard form)}
- \left(x - 6\right) \left(13 x - 40\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = \frac{40}{13},\quad x = 6
Check: each root substituted back into the equation gives 0.
Answer: x = \frac{40}{13},\quad x = 6
Question 1(xiv)
Solve, each of the following equations, using the formula : \frac{x-1}{x-2} + \frac{x-3}{x-4} = 3\frac{1}{3}
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
- 4 x^{2} + 30 x - 50 = 0 \quad\text{(standard form)}
- 2 \left(x - 5\right) \left(2 x - 5\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = \frac{5}{2},\quad x = 5
Check: each root substituted back into the equation gives 0.
Answer: x = \frac{5}{2},\quad x = 5
Question 2(i)
Solve each of the following equations for x and give, in each case, your answer correct to one decimal place : x^2 - 8x + 5 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
x^{2} - 8 x + 5 = 0 \quad\text{(standard form)}
a=1,\; b=-8,\; c=5,\quad D=b^2-4ac=44
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = 0.7,\quad x = 7.3
Check: each root substituted back into the equation gives 0.
Answer: x = 0.7,\quad x = 7.3 (rounded to 1 decimal place)
Question 2(ii)
Solve each of the following equations for x and give, in each case, your answer correct to one decimal place : 5x^2 + 10x - 3 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
5 x^{2} + 10 x - 3 = 0 \quad\text{(standard form)}
a=5,\; b=10,\; c=-3,\quad D=b^2-4ac=160
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = 0.3,\quad x = -2.3
Check: each root substituted back into the equation gives 0.
Answer: x = 0.3,\quad x = -2.3 (rounded to 1 decimal place)
Question 3(i)
Solve each of the following equations for x and give, in each case, your answer correct to 2 decimal places : 2x^2 - 10x + 5 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
2 x^{2} - 10 x + 5 = 0 \quad\text{(standard form)}
a=2,\; b=-10,\; c=5,\quad D=b^2-4ac=60
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = 0.56,\quad x = 4.44
Check: each root substituted back into the equation gives 0.
Answer: x = 0.56,\quad x = 4.44 (rounded to 2 decimal places)
Question 3(ii)
Solve each of the following equations for x and give, in each case, your answer correct to 2 decimal places : 4x + \frac{6}{x} + 13 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
4 x^{2} + 13 x + 6 = 0 \quad\text{(standard form)}
a=4,\; b=13,\; c=6,\quad D=b^2-4ac=73
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = -2.69,\quad x = -0.56
Check: each root substituted back into the equation gives 0.
Answer: x = -2.69,\quad x = -0.56 (rounded to 2 decimal places)
Question 3(iii)
Solve each of the following equations for x and give, in each case, your answer correct to 2 decimal places : x^2 - 3x - 9 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
x^{2} - 3 x - 9 = 0 \quad\text{(standard form)}
a=1,\; b=-3,\; c=-9,\quad D=b^2-4ac=45
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = -1.85,\quad x = 4.85
Check: each root substituted back into the equation gives 0.
Answer: x = -1.85,\quad x = 4.85 (rounded to 2 decimal places)
Question 3(iv)
Solve each of the following equations for x and give, in each case, your answer correct to 2 decimal places : x^2 - 5x - 10 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
x^{2} - 5 x - 10 = 0 \quad\text{(standard form)}
a=1,\; b=-5,\; c=-10,\quad D=b^2-4ac=65
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = -1.53,\quad x = 6.53
Check: each root substituted back into the equation gives 0.
Answer: x = -1.53,\quad x = 6.53 (rounded to 2 decimal places)
Question 4(i)
Solve each of the following equations for x, giving your answer correct to 3 decimal places: 3x^2 - 12x - 1 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
3 x^{2} - 12 x - 1 = 0 \quad\text{(standard form)}
a=3,\; b=-12,\; c=-1,\quad D=b^2-4ac=156
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = -0.082,\quad x = 4.082
Check: each root substituted back into the equation gives 0.
Answer: x = -0.082,\quad x = 4.082 (rounded to 3 decimal places)
Question 4(ii)
Solve each of the following equations for x, giving your answer correct to 3 decimal places: x^2 - 16x + 6 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
x^{2} - 16 x + 6 = 0 \quad\text{(standard form)}
a=1,\; b=-16,\; c=6,\quad D=b^2-4ac=232
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = 0.384,\quad x = 15.616
Check: each root substituted back into the equation gives 0.
Answer: x = 0.384,\quad x = 15.616 (rounded to 3 decimal places)
Question 4(iii)
Solve each of the following equations for x, giving your answer correct to 3 decimal places: 2x^2 + 11x + 4 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
2 x^{2} + 11 x + 4 = 0 \quad\text{(standard form)}
a=2,\; b=11,\; c=4,\quad D=b^2-4ac=89
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = -5.108,\quad x = -0.392
Check: each root substituted back into the equation gives 0.
Answer: x = -5.108,\quad x = -0.392 (rounded to 3 decimal places)
Question 5(i)
Solve : x^4 - 2x^2 - 3 = 0
Method: This is a biquadratic: substitute y=x^2 to get a quadratic in y, solve it, then back-substitute x^2=y.
Working:
x^{4} - 2 x^{2} - 3 = 0
\text{Let } x^2 = y: \quad y^{2} - 2 y - 3 = 0
\Rightarrow\; y = -1,\quad y = 3
\text{Back-substitute } x^2 = y:
x = - \sqrt{3},\quad x = \sqrt{3}
Check: each root substituted into the quartic gives 0.
Answer: x = - \sqrt{3},\quad x = \sqrt{3}
Question 5(ii)
Solve : x^4 - 10x^2 + 9 = 0
Method: This is a biquadratic: substitute y=x^2 to get a quadratic in y, solve it, then back-substitute x^2=y.
Working:
x^{4} - 10 x^{2} + 9 = 0
\text{Let } x^2 = y: \quad y^{2} - 10 y + 9 = 0
\Rightarrow\; y = 1,\quad y = 9
\text{Back-substitute } x^2 = y:
x = -1,\quad x = 1,\quad x = -3,\quad x = 3
Check: each root substituted into the quartic gives 0.
Answer: x = -1,\quad x = 1,\quad x = -3,\quad x = 3
Question 6(i)
Solve : (x^2 - x)^2 + 5(x^2 - x) + 4 = 0
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{4} - 2 x^{3} + 6 x^{2} - 5 x + 4 = 0
D = b^2 - 4ac = -71 \lt 0
Check: discriminant < 0, so there is no real solution.
Answer: no real roots
Question 6(ii)
Solve : (x^2 - 3x)^2 - 16(x^2 - 3x) - 36 = 0
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{4} - 6 x^{3} - 7 x^{2} + 48 x - 36 = 0 \quad\text{(standard form)}
\left(x - 6\right) \left(x - 2\right) \left(x - 1\right) \left(x + 3\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -3,\quad x = 1,\quad x = 2,\quad x = 6
Check: each root substituted back into the equation gives 0.
Answer: x = -3,\quad x = 1,\quad x = 2,\quad x = 6
Question 7(i)
Solve : \sqrt{\frac{x}{x-3}} + \sqrt{\frac{x-3}{x}} = \frac{5}{2}
Method: We substitute t for the square-root term (its reciprocal is \frac1t), solve t+\frac1t=k, then back-substitute and reject any extraneous value.
Working:
\text{Let } t = \sqrt{\frac{x}{x - 3}} \;\Rightarrow\; \sqrt{\frac{x - 3}{x}} = \dfrac{1}{t}
t + \dfrac{1}{t} = \frac{5}{2} \;\Rightarrow\; t^2 - \frac{5}{2}\,t + 1 = 0
t = \frac{1}{2},\; 2 \quad\text{(both positive)}
\text{Back-substitute } \frac{x}{x - 3} = t^2:
x = -1,\quad x = 4
Check: each x substituted into the original radical equation gives \frac{5}{2} (radicands real and non-negative).
Answer: x = -1,\quad x = 4
Question 7(ii)
Solve : \left(\frac{2x-3}{x-1}\right) - 4\left(\frac{x-1}{2x-3}\right) = 3
Method: We reduce the equation to standard form and solve for every root.
Working:
- 6 x^{2} + 11 x - 4 = 0 \quad\text{(standard form)}
- \left(2 x - 1\right) \left(3 x - 4\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = \frac{1}{2},\quad x = \frac{4}{3}
Check: each root substituted back into the equation gives 0.
Answer: x = \frac{1}{2},\quad x = \frac{4}{3}
Question 7(iii)
Solve : \left(\frac{3x+1}{x+1}\right) + \left(\frac{x+1}{3x+1}\right) = \frac{5}{2}
Method: We reduce the equation to standard form and solve for every root.
Working:
5 x^{2} - 4 x - 1 = 0 \quad\text{(standard form)}
\left(x - 1\right) \left(5 x + 1\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = - \frac{1}{5},\quad x = 1
Check: each root substituted back into the equation gives 0.
Answer: x = - \frac{1}{5},\quad x = 1
Question 8
Solve the equation 2x - \frac{1}{x} = 7. Write your answer correct to two decimal places.
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
2 x^{2} - 7 x - 1 = 0 \quad\text{(standard form)}
a=2,\; b=-7,\; c=-1,\quad D=b^2-4ac=57
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = -0.14,\quad x = 3.64
Check: each root substituted back into the equation gives 0.
Answer: x = -0.14,\quad x = 3.64 (rounded to 2 decimal places)
Question 9
Solve the following equation and give your answer correct to 3 significant figures : 5x^2 - 3x - 4 = 0
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
5 x^{2} - 3 x - 4 = 0 \quad\text{(standard form)}
a=5,\; b=-3,\; c=-4,\quad D=b^2-4ac=89
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = -0.643,\quad x = 1.24
Check: each root substituted back into the equation gives 0.
Answer: x = -0.643,\quad x = 1.24 (to 3 significant figures)
Question 10
Solve for x using the quadratic formula. Write your answer correct to two significant figures. (x - 1)^2 - 3x + 4 = 0.
Method: We use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} and give the answer to the required accuracy.
Working:
x^{2} - 5 x + 5 = 0 \quad\text{(standard form)}
a=1,\; b=-5,\; c=5,\quad D=b^2-4ac=5
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = 1.4,\quad x = 3.6
Check: each root substituted back into the equation gives 0.
Answer: x = 1.4,\quad x = 3.6 (to 2 significant figures)
Exercise 5(D)
Solve a mixed set of quadratic equations by choosing the most efficient method for each problem: use factorisation when the coefficients factor neatly, and fall back on the quadratic formula when they do not.
Question 1
\frac{2x}{x - 3} +\frac{1}{2x + 3} +\frac{3x + 9}{(x - 3)(2x + 3)} = 0; x \neq 3, x \neq - \frac {3}{2}
Method: We reduce the equation to standard form and solve for every root.
Working:
4 x^{2} + 10 x + 6 = 0 \quad\text{(standard form)}
2 \left(x + 1\right) \left(2 x + 3\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = - \frac{3}{2},\quad x = -1
Check: each root substituted back into the equation gives 0.
Answer: x = - \frac{3}{2},\quad x = -1
Question 2
Method: We reduce the equation to standard form and solve for every root.
Working:
4 x^{2} + 12 x - 72 = 0 \quad\text{(standard form)}
4 \left(x - 3\right) \left(x + 6\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -6,\quad x = 3
Check: each root substituted back into the equation gives 0.
Answer: x = -6,\quad x = 3
Question 3
Method: We reduce the equation to standard form and solve for every root.
Working:
a^{2} x^{2} - b^{2} = 0 \quad\text{(standard form)}
a=a^{2},\; b=0,\; c=- b^{2},\quad D=b^2-4ac=4 a^{2} b^{2}
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = - \frac{b}{a},\quad x = \frac{b}{a}
Check: each root substituted back into the equation gives 0.
Answer: x = - \frac{b}{a},\quad x = \frac{b}{a}
Question 4
Method: We reduce the equation to standard form and solve for every root.
Working:
8 x^{2} - 22 x + 15 = 0 \quad\text{(standard form)}
\left(2 x - 3\right) \left(4 x - 5\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = \frac{5}{4},\quad x = \frac{3}{2}
Check: each root substituted back into the equation gives 0.
Answer: x = \frac{5}{4},\quad x = \frac{3}{2}
Question 5
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{2} + 4 x + 4 = 0 \quad\text{(standard form)}
\left(x + 2\right)^{2} = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -2
Check: each root substituted back into the equation gives 0.
Answer: x = -2
Question 6
Method: This is a biquadratic: substitute y=x^2 to get a quadratic in y, solve it, then back-substitute x^2=y.
Working:
2 x^{4} - 5 x^{2} + 3 = 0
\text{Let } x^2 = y: \quad 2 y^{2} - 5 y + 3 = 0
\Rightarrow\; y = 1,\quad y = \frac{3}{2}
\text{Back-substitute } x^2 = y:
x = -1,\quad x = 1,\quad x = - \frac{\sqrt{6}}{2},\quad x = \frac{\sqrt{6}}{2}
Check: each root substituted into the quartic gives 0.
Answer: x = -1,\quad x = 1,\quad x = - \frac{\sqrt{6}}{2},\quad x = \frac{\sqrt{6}}{2}
Question 7
Method: This is a biquadratic: substitute y=x^2 to get a quadratic in y, solve it, then back-substitute x^2=y.
Working:
x^{4} - 2 x^{2} - 3 = 0
\text{Let } x^2 = y: \quad y^{2} - 2 y - 3 = 0
\Rightarrow\; y = -1,\quad y = 3
\text{Back-substitute } x^2 = y:
x = - \sqrt{3},\quad x = \sqrt{3}
Check: each root substituted into the quartic gives 0.
Answer: x = - \sqrt{3},\quad x = \sqrt{3}
Question 8
Method: This is a reciprocal equation: substitute y=x\pm\frac1x (so x^2+\frac1{x^2}=y^2\mp2), solve the quadratic in y, then back-substitute.
Working:
\text{Let } y = x + \frac{1}{x} \;\Rightarrow\; x^2 + \frac{1}{x^2} = y^2 - 2
\text{The equation reduces to } 9 y^{2} - 9 y - 70 = 0
\Rightarrow\; y = - \frac{7}{3},\quad y = \frac{10}{3}
\text{Back-substitute } x + \frac{1}{x} = y:
x = - \frac{7}{6} - \frac{\sqrt{13}}{6},\quad x = - \frac{7}{6} + \frac{\sqrt{13}}{6},\quad x = \frac{1}{3},\quad x = 3
Check: each root substituted into the original equation gives 0.
Answer: x = - \frac{7}{6} - \frac{\sqrt{13}}{6},\quad x = - \frac{7}{6} + \frac{\sqrt{13}}{6},\quad x = \frac{1}{3},\quad x = 3
Question 9
Method: This is a reciprocal equation: substitute y=x\pm\frac1x (so x^2+\frac1{x^2}=y^2\mp2), solve the quadratic in y, then back-substitute.
Working:
\text{Let } y = x + \frac{1}{x} \;\Rightarrow\; x^2 + \frac{1}{x^2} = y^2 - 2
\text{The equation reduces to } 2 y^{2} - y - 15 = 0
\Rightarrow\; y = - \frac{5}{2},\quad y = 3
\text{Back-substitute } x + \frac{1}{x} = y:
x = -2,\quad x = - \frac{1}{2},\quad x = \frac{3}{2} - \frac{\sqrt{5}}{2},\quad x = \frac{\sqrt{5}}{2} + \frac{3}{2}
Check: each root substituted into the original equation gives 0.
Answer: x = -2,\quad x = - \frac{1}{2},\quad x = \frac{3}{2} - \frac{\sqrt{5}}{2},\quad x = \frac{\sqrt{5}}{2} + \frac{3}{2}
Question 10
Method: This is a reciprocal equation: substitute y=x\pm\frac1x (so x^2+\frac1{x^2}=y^2\mp2), solve the quadratic in y, then back-substitute.
Working:
\text{Let } y = x - \frac{1}{x} \;\Rightarrow\; x^2 + \frac{1}{x^2} = y^2 + 2
\text{The equation reduces to } y^{2} - 3 y = 0
\Rightarrow\; y = 0,\quad y = 3
\text{Back-substitute } x - \frac{1}{x} = y:
x = -1,\quad x = 1,\quad x = \frac{3}{2} - \frac{\sqrt{13}}{2},\quad x = \frac{3}{2} + \frac{\sqrt{13}}{2}
Check: each root substituted into the original equation gives 0.
Answer: x = -1,\quad x = 1,\quad x = \frac{3}{2} - \frac{\sqrt{13}}{2},\quad x = \frac{3}{2} + \frac{\sqrt{13}}{2}
Question 11
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{4} + 10 x^{3} + 35 x^{2} + 50 x - 96 = 0 \quad\text{(standard form)}
\text{(Textbook hint: use the substitution } \[ {Take} x ^ {2} + 5 x = y\text{)}
\left(x - 1\right) \left(x + 6\right) \left(x^{2} + 5 x + 16\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -6,\quad x = 1
Check: each root substituted back into the equation gives 0.
Answer: x = -6,\quad x = 1
Question 12(i)
Solve each of the following equations, giving answer up to two decimal places. x^{2} - 5x - 10 = 0
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{2} - 5 x - 10 = 0 \quad\text{(standard form)}
a=1,\; b=-5,\; c=-10,\quad D=b^2-4ac=65
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = \frac{5}{2} - \frac{\sqrt{65}}{2},\quad x = \frac{5}{2} + \frac{\sqrt{65}}{2}
Check: each root substituted back into the equation gives 0.
Answer: x = \frac{5}{2} - \frac{\sqrt{65}}{2},\quad x = \frac{5}{2} + \frac{\sqrt{65}}{2}
Question 12(ii)
Solve each of the following equations, giving answer up to two decimal places. 3x^{2} - x - 7 = 0
Method: We reduce the equation to standard form and solve for every root.
Working:
3 x^{2} - x - 7 = 0 \quad\text{(standard form)}
a=3,\; b=-1,\; c=-7,\quad D=b^2-4ac=85
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = \frac{1}{6} - \frac{\sqrt{85}}{6},\quad x = \frac{1}{6} + \frac{\sqrt{85}}{6}
Check: each root substituted back into the equation gives 0.
Answer: x = \frac{1}{6} - \frac{\sqrt{85}}{6},\quad x = \frac{1}{6} + \frac{\sqrt{85}}{6}
Question 13
Solve: \left(\frac{x}{x + 2}\right)^2 - 7\left(\frac{x}{x + 2}\right) + 12 = 0; x \neq -2.
Method: We reduce the equation to standard form and solve for every root.
Working:
6 x^{2} + 34 x + 48 = 0 \quad\text{(standard form)}
2 \left(x + 3\right) \left(3 x + 8\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -3,\quad x = - \frac{8}{3}
Check: each root substituted back into the equation gives 0.
Answer: x = -3,\quad x = - \frac{8}{3}
Question 14(i)
Solve : x^{2} - 11x - 12 = 0 ; when x \in \mathbb{N}
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{2} - 11 x - 12 = 0 \quad\text{(standard form)}
\left(x - 12\right) \left(x + 1\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -1,\quad x = 12
Check: each root substituted back into the equation gives 0.
Answer: x = -1,\quad x = 12
Question 14(ii)
Solve : x^{2} - 4x - 12 = 0 ; when x \in I
Method: We reduce the equation to standard form and solve for every root.
Working:
x^{2} - 4 x - 12 = 0 \quad\text{(standard form)}
\left(x - 6\right) \left(x + 2\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -2,\quad x = 6
Check: each root substituted back into the equation gives 0.
Answer: x = -2,\quad x = 6
Question 14(iii)
Solve : 2x^{2} - 9x + 10 = 0 ; when x \in \mathbb{Q} .
Method: We reduce the equation to standard form and solve for every root.
Working:
2 x^{2} - 9 x + 10 = 0 \quad\text{(standard form)}
\left(x - 2\right) \left(2 x - 5\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = 2,\quad x = \frac{5}{2}
Check: each root substituted back into the equation gives 0.
Answer: x = 2,\quad x = \frac{5}{2}
Question 15
Method: We reduce the equation to standard form and solve for every root.
Working:
a^{2} x^{2} + 2 a b x^{2} - a x + b^{2} x^{2} - b x - 6 = 0 \quad\text{(standard form)}
\text{(Textbook hint: use the substitution } (a + b) x = y\text{)}
a=a^{2} + 2 a b + b^{2},\; b=- a - b,\; c=-6,\quad D=b^2-4ac=25 a^{2} + 50 a b + 25 b^{2}
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = - \frac{2}{a + b},\quad x = \frac{3}{a + b}
Check: each root substituted back into the equation gives 0.
Answer: x = - \frac{2}{a + b},\quad x = \frac{3}{a + b}
Question 16
Method: We reduce the equation to standard form and solve for every root.
Working:
p^{2} q + p^{2} x + p q^{2} + 2 p q x + p x^{2} + q^{2} x + q x^{2} = 0 \quad\text{(standard form)}
\text{(Textbook hint: use the substitution } {Take}: \left(\frac {1}{p} + \frac {1}{q}\right) + \left(\text{)} \]
a=p + q,\; b=p^{2} + 2 p q + q^{2},\; c=p q \left(p + q\right),\quad D=b^2-4ac=p^{4} - 2 p^{2} q^{2} + q^{4}
x = \dfrac{-b \pm \sqrt{D}}{2a}
x = - p,\quad x = - q
Check: each root substituted back into the equation gives 0.
Answer: x = - p,\quad x = - q
Question 17(i)
Solve : x(x + 1) + (x + 2)(x + 3) = 42
Method: We reduce the equation to standard form and solve for every root.
Working:
2 x^{2} + 6 x - 36 = 0 \quad\text{(standard form)}
2 \left(x - 3\right) \left(x + 6\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = -6,\quad x = 3
Check: each root substituted back into the equation gives 0.
Answer: x = -6,\quad x = 3
Question 17(ii)
Solve : \frac{1}{x + 1} -\frac{2}{x + 2} = \frac{3}{x + 3} -\frac{4}{x + 4}
Method: We reduce the equation to standard form and solve for every root.
Working:
- 4 x^{2} - 10 x = 0 \quad\text{(standard form)}
- 2 x \left(2 x + 5\right) = 0 \quad\text{(factorising)}
\text{By the zero-product rule, each factor } = 0:
x = - \frac{5}{2},\quad x = 0
Check: each root substituted back into the equation gives 0.
Answer: x = - \frac{5}{2},\quad x = 0
Question 18(i)
For each equation, given below, find the value of ‘m’ so that the equation has equal roots. Also, find the solution of each equation : (m - 3)x^2 - 4x + 1 = 0
Method: “Equal roots” means the discriminant is zero, so we set D=b^2-4ac=0 and solve for the unknown constant (and, where asked, the equal root itself).
Working:
a = m - 3,\; b = -4,\; c = 1
D = b^2 - 4ac = 28 - 4 m
\text{For the required roots, } D = 0
\text{At } m=7:\; 4 x^{2} - 4 x + 1=0\Rightarrow x=\frac{1}{2}
Check: substituting the value back gives D = 0 and each root checks out.
Answer: m = 7\;\Rightarrow\; x = \frac{1}{2}
Question 18(ii)
For each equation, given below, find the value of ‘m’ so that the equation has equal roots. Also, find the solution of each equation : 3x^2 + 12x + (m + 7) = 0
Method: “Equal roots” means the discriminant is zero, so we set D=b^2-4ac=0 and solve for the unknown constant (and, where asked, the equal root itself).
Working:
a = 3,\; b = 12,\; c = m + 7
D = b^2 - 4ac = 60 - 12 m
\text{For the required roots, } D = 0
\text{At } m=5:\; 3 x^{2} + 12 x + 12=0\Rightarrow x=-2
Check: substituting the value back gives D = 0 and each root checks out.
Answer: m = 5\;\Rightarrow\; x = -2
Question 18(iii)
For each equation, given below, find the value of ‘m’ so that the equation has equal roots. Also, find the solution of each equation : x^2 - (m + 2)x + (m + 5) = 0
Method: “Equal roots” means the discriminant is zero, so we set D=b^2-4ac=0 and solve for the unknown constant (and, where asked, the equal root itself).
Working:
a = 1,\; b = - m - 2,\; c = m + 5
D = b^2 - 4ac = m^{2} - 16
\text{For the required roots, } D = 0
\text{At } m=-4:\; x^{2} + 2 x + 1=0\Rightarrow x=-1
\text{At } m=4:\; x^{2} - 6 x + 9=0\Rightarrow x=3
Check: substituting the value back gives D = 0 and each root checks out.
Answer: m = -4\;\Rightarrow\; x = -1,\quad m = 4\;\Rightarrow\; x = 3
Question 19
Without solving the following quadratic equation, find the value of ‘p’ for which the roots are equal. px^2 - 4x + 3 = 0 \quad
Method: We judge the nature of the roots without solving, using the discriminant D=b^2-4ac: D \gt 0 gives two distinct real roots, D=0 two equal real roots, and D \lt 0 means the roots are not real.
Working:
a = p,\; b = -4,\; c = 3
D = b^2 - 4ac = 16 - 12 p
Check: D = 16 – 12 p.
Answer: depends on the sign of the discriminant D = 16 - 12 p
Question 20
Without solving the following quadratic equation, find the value of ‘m’ for which the given equation has real and equal roots. x^2 + 2(m - 1)x + (m + 5) = 0 \quad
Method: We judge the nature of the roots without solving, using the discriminant D=b^2-4ac: D \gt 0 gives two distinct real roots, D=0 two equal real roots, and D \lt 0 means the roots are not real.
Working:
a = 1,\; b = 2 m - 2,\; c = m + 5
D = b^2 - 4ac = 4 m^{2} - 12 m - 16
Check: D = 4 m^{2} – 12 m – 16.
Answer: depends on the sign of the discriminant D = 4 m^{2} - 12 m - 16
Common mistakes to avoid
- Forgetting to rewrite the equation in the standard form ax^2 + bx + c = 0 before identifying a, b, and c, which leads to incorrect signs and values.
- Making sign errors when splitting the middle term, particularly when the product ac is negative and one component must be negative while the other is positive.
- Incorrectly calculating the discriminant by computing b^2 + 4ac instead of b^2 - 4ac, which completely changes the nature of the roots.
- Omitting the \pm sign in the quadratic formula, which results in only one root being found instead of the required two distinct roots.
Exam tips
- Always write the equation in standard form ax^2 + bx + c = 0 as your first step, and explicitly state the values of a, b, and c to avoid careless errors.
- When a question asks about the nature of the roots, show the calculation of the discriminant D on its own line and state the condition it satisfies before drawing your conclusion.
- If factorisation does not yield integer factors within a reasonable time, switch immediately to the quadratic formula rather than forcing a factorisation that may not exist.
Frequently asked questions
How do I decide whether to use factorisation or the quadratic formula for a given problem?
Try factorisation first if the coefficients are small and the product ac splits neatly into two numbers that add to b. If the numbers are large, involve decimals, or do not split cleanly, use the quadratic formula to save time and reduce the risk of arithmetic errors.
What does it mean when the discriminant is a perfect square?
If the discriminant D = b^2 - 4ac is a perfect square (and non-negative), the roots are real and rational. If it is positive but not a perfect square, the roots are real but irrational, involving surds in the final answer.
Why is it necessary that a \neq 0 in a quadratic equation?
If a = 0, the x^2 term vanishes and the equation reduces to a linear equation bx + c = 0, which has only one root. The term ‘quadratic’ specifically requires the highest power of the variable to be two, so a must never be zero.