ICSE Class 10 Maths Quadratic Equations: direct answer
ICSE Class 10 Maths quadratic equations are second degree equations that can be written in the form ax^2+bx+c=0, where a \ne 0. In Concise Mathematics Selina Class 10 Chapter 5, the main skill is to choose the right method: factorisation when clean factors are visible, completing the square when the equation is suited to it, the quadratic formula when factorisation is not quick, and the discriminant to decide the nature of roots.
This replacement page is written as a study-and-solutions page for Selina Chapter 5. Since exercise numbering and question counts can vary by print edition, the worked answers below use the question types visible on this URL and the standard CISCE Class 10 treatment of quadratic equations. Each model solution shows the working in school format so that a student can check the method, not only the final answer.
For nearby study pages, use the ICSE Class 10 solutions, the ICSE textbook solutions index, and the ICSE Class 10 study resources. For official syllabus and examination publications, refer to the CISCE official website.
Chapter 5 formulas and methods for Concise Mathematics Selina Solutions Class 10 ICSE Chapter 5 Quadratic Equations
Before solving any question, first bring the equation to standard form. This prevents sign errors and makes the method clear.
| Use case | Method | What to check |
|---|---|---|
| Equation already splits into two factors | Use the zero product rule: if AB=0, then A=0 or B=0. | Write both roots. Do not give only one root. |
| Equation has a square or identity | Use identities such as \(a^2-b^2=(a+b)(a-b)\). | After taking square root, include both + and - possibilities. |
| Equation has fractions | State restrictions, clear denominators, then solve the quadratic. | Reject any value that makes a denominator zero. |
| Factorisation is not direct | Use x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}. | Substitute a, b, and c with correct signs. |
| Question asks nature of roots | Use the discriminant D=b^2-4ac. | D>0 means two distinct real roots, D=0 means equal real roots, and D<0 means no real roots. |
Concept snapshot for quadratic roots
Think of a quadratic equation as a curve that may cut the x-axis at two points, touch it at one point, or miss it. The roots are the x-values where the curve y=ax^2+bx+c meets the x-axis. This is why a quadratic can have two real roots, one repeated real root, or no real root. Algebraically, the discriminant D=b^2-4ac tells the same story before you solve the equation.
Exercise 5(A): roots and substitution type questions
These questions test whether you understand what a root means. A number is a root of a quadratic equation only when it satisfies the equation after substitution.
Question 1(a): Solve 4x^2-9=0
Step 1: Move the constant term to the other side.
4x^2=9
Step 2: Divide both sides by 4.
x^2=\frac{9}{4}
Step 3: Take square roots on both sides. A square root step gives two values.
x=\pm\sqrt{\frac{9}{4}}=\pm\frac{3}{2}
Final answer: x=\dfrac{3}{2} or x=-\dfrac{3}{2}.
Question 1(b): Solve \((x-3)(x+5)=0\)
Step 1: Use the zero product rule.
x-3=0 \quad \text{or} \quad x+5=0
Step 2: Solve the two linear equations separately.
x=3 \quad \text{or} \quad x=-5
Final answer: x=3 or x=-5.
Question 1(c): If 4 is a root of x^2+kx-4=0, find k
Step 1: Since 4 is a root, substitute x=4 in the equation.
(4)^2+k(4)-4=0
Step 2: Simplify and solve for k.
16+4k-4=0
4k+12=0
4k=-12
k=-3
Final answer: k=-3.
Question 2: If \sqrt{\dfrac{2}{3}} is a solution of 3x^2+mx+2=0, find m
Step 1: Substitute x=\sqrt{\dfrac{2}{3}}.
3\left(\sqrt{\frac{2}{3}}\right)^2+m\sqrt{\frac{2}{3}}+2=0
Step 2: Square the surd and simplify.
3\left(\frac{2}{3}\right)+m\sqrt{\frac{2}{3}}+2=0
2+m\sqrt{\frac{2}{3}}+2=0
m\sqrt{\frac{2}{3}}=-4
Step 3: Divide by \sqrt{\dfrac{2}{3}} and simplify the surd.
m=-4\div\sqrt{\frac{2}{3}}
m=-4\sqrt{\frac{3}{2}}=-2\sqrt{6}
Final answer: m=-2\sqrt{6}.
Question 3: If \dfrac{2}{3} and 1 are roots of mx^2+nx+6=0, find m and n
Step 1: For ax^2+bx+c=0, product of roots is \dfrac{c}{a}. Here the product is \dfrac{2}{3}\times1=\dfrac{2}{3}.
\frac{6}{m}=\frac{2}{3}
18=2m
m=9
Step 2: Sum of roots is -\dfrac{b}{a}. Here the sum is \dfrac{2}{3}+1=\dfrac{5}{3}.
-\frac{n}{m}=\frac{5}{3}
Step 3: Substitute m=9.
-\frac{n}{9}=\frac{5}{3}
-n=15
n=-15
Final answer: m=9 and n=-15.
Exercise 5(B): factorisation type questions
In factorisation questions, the aim is to convert the quadratic expression into a product of two linear factors. Always check that the two numbers used for splitting the middle term add to b and multiply to ac.
Question 1(a): Find the roots of x^2-6x-7=0
Step 1: Split the middle term. The numbers -7 and 1 add to -6 and multiply to -7.
x^2-7x+x-7=0
Step 2: Group the terms and factorise.
x(x-7)+1(x-7)=0
(x+1)(x-7)=0
Step 3: Apply the zero product rule.
x+1=0 \quad \text{or} \quad x-7=0
Final answer: x=-1 or x=7.
Question 2: Solve \((2x-3)^2=49\) by factorisation
Step 1: Write 49 as 7^2 and bring all terms to one side.
(2x-3)^2-7^2=0
Step 2: Use \(a^2-b^2=(a+b)(a-b)\).
(2x-3+7)(2x-3-7)=0
(2x+4)(2x-10)=0
Step 3: Solve each factor.
2x+4=0 \quad \text{or} \quad 2x-10=0
x=-2 \quad \text{or} \quad x=5
Final answer: x=-2 or x=5.
Question 3: Solve \((x+1)(2x+8)=(x+7)(x+3)\)
Step 1: Expand both sides.
(x+1)(2x+8)=2x^2+10x+8
(x+7)(x+3)=x^2+10x+21
Step 2: Equate and bring all terms to one side.
2x^2+10x+8=x^2+10x+21
x^2-13=0
Step 3: Solve by difference of squares over surds.
x^2=13
x=\pm\sqrt{13}
Final answer: x=\sqrt{13} or x=-\sqrt{13}.
Question 4: Solve \(4(2x-3)^2-(2x-3)-14=0\)
Step 1: Put y=2x-3. Then the equation becomes:
4y^2-y-14=0
Step 2: Factorise the quadratic in y.
4y^2-8y+7y-14=0
4y(y-2)+7(y-2)=0
(4y+7)(y-2)=0
Step 3: Hence, y=-\dfrac{7}{4} or y=2.
Step 4: Substitute back y=2x-3.
2x-3=-\frac{7}{4}\quad \Rightarrow \quad 2x=\frac{5}{4}\quad \Rightarrow \quad x=\frac{5}{8}
2x-3=2\quad \Rightarrow \quad 2x=5\quad \Rightarrow \quad x=\frac{5}{2}
Final answer: x=\dfrac{5}{8} or x=\dfrac{5}{2}.
Question 5: Solve 2x^2-9x+10=0 when x\in\mathbb{N} and when x\in\mathbb{Q}
Step 1: Factorise the equation.
2x^2-9x+10=0
2x^2-4x-5x+10=0
2x(x-2)-5(x-2)=0
(2x-5)(x-2)=0
Step 2: Find the roots.
2x-5=0 \quad \text{or} \quad x-2=0
x=\frac{5}{2} \quad \text{or} \quad x=2
Step 3: Apply the given set condition. Assuming \mathbb{N}=\{1,2,3,\ldots\}, only 2 is natural. Both \dfrac{5}{2} and 2 are rational.
Final answer: If x\in\mathbb{N}, x=2. If x\in\mathbb{Q}, x=\dfrac{5}{2} or x=2.
Question 6: Solve \dfrac{x-3}{x+3}+\dfrac{x+3}{x-3}=2\dfrac{1}{2}
Step 1: State restrictions: x\ne -3 and x\ne 3.
Step 2: Convert the mixed number and combine the fractions.
\frac{x-3}{x+3}+\frac{x+3}{x-3}=\frac{5}{2}
\frac{(x-3)^2+(x+3)^2}{(x+3)(x-3)}=\frac{5}{2}
Step 3: Expand and simplify.
\frac{x^2-6x+9+x^2+6x+9}{x^2-9}=\frac{5}{2}
\frac{2x^2+18}{x^2-9}=\frac{5}{2}
Step 4: Cross multiply.
2(2x^2+18)=5(x^2-9)
4x^2+36=5x^2-45
x^2=81
x=\pm 9
Step 5: Both 9 and -9 are allowed because neither makes a denominator zero.
Final answer: x=9 or x=-9.
Question 7: Solve \dfrac{4}{x+2}-\dfrac{1}{x+3}=\dfrac{4}{2x+1}
Step 1: Restrictions are x\ne -2, x\ne -3, and x\ne -\dfrac{1}{2}.
Step 2: Combine the left side.
\frac{4(x+3)-(x+2)}{(x+2)(x+3)}=\frac{4}{2x+1}
\frac{3x+10}{(x+2)(x+3)}=\frac{4}{2x+1}
Step 3: Cross multiply.
(3x+10)(2x+1)=4(x+2)(x+3)
6x^2+23x+10=4x^2+20x+24
2x^2+3x-14=0
Step 4: Factorise.
2x^2+7x-4x-14=0
x(2x+7)-2(2x+7)=0
(x-2)(2x+7)=0
Final answer: x=2 or x=-\dfrac{7}{2}. Both satisfy the restrictions.
Question 8: Solve \dfrac{5}{x-2}-\dfrac{3}{x+6}=\dfrac{4}{x}
Step 1: Restrictions are x\ne 2, x\ne -6, and x\ne 0.
Step 2: Combine the left side.
\frac{5(x+6)-3(x-2)}{(x-2)(x+6)}=\frac{4}{x}
\frac{2x+36}{(x-2)(x+6)}=\frac{4}{x}
Step 3: Cross multiply and simplify.
x(2x+36)=4(x-2)(x+6)
2x^2+36x=4(x^2+4x-12)
2x^2+36x=4x^2+16x-48
x^2-10x-24=0
Step 4: Factorise.
x^2-12x+2x-24=0
x(x-12)+2(x-12)=0
(x+2)(x-12)=0
Final answer: x=-2 or x=12. Both values are valid.
Question 9: Solve \(\left(1+\dfrac{1}{x+1}\right)\left(1-\dfrac{1}{x-1}\right)=\dfrac{7}{8}\)
Step 1: Restrictions are x\ne -1 and x\ne 1.
Step 2: Simplify each bracket.
1+\frac{1}{x+1}=\frac{x+2}{x+1}
1-\frac{1}{x-1}=\frac{x-2}{x-1}
Step 3: Substitute and cross multiply.
\frac{(x+2)(x-2)}{(x+1)(x-1)}=\frac{7}{8}
\frac{x^2-4}{x^2-1}=\frac{7}{8}
8(x^2-4)=7(x^2-1)
8x^2-32=7x^2-7
x^2=25
x=\pm 5
Final answer: x=5 or x=-5. Both values are valid.
Completing the square and quadratic formula
Use completing the square when the equation has a clear x^2+2px pattern. Use the quadratic formula when clean factorisation is not visible. Both methods are part of the standard treatment of quadratic equations in ICSE Class 10 Maths.
Worked example: Solve x^2-6x+2=0 by completing the square
Step 1: Move the constant term to the other side.
x^2-6x=-2
Step 2: Half of 6 is 3, and 3^2=9. Add 9 to both sides.
x^2-6x+9=-2+9
(x-3)^2=7
Step 3: Take square roots.
x-3=\pm\sqrt{7}
Final answer: x=3+\sqrt{7} or x=3-\sqrt{7}.
Worked example: Solve 2x^2-4x-3=0 by the quadratic formula
Step 1: Identify a=2, b=-4, and c=-3.
Step 2: Use the quadratic formula.
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
x=\frac{-(-4)\pm\sqrt{(-4)^2-4(2)(-3)}}{2(2)}
x=\frac{4\pm\sqrt{16+24}}{4}
x=\frac{4\pm\sqrt{40}}{4}
x=\frac{4\pm 2\sqrt{10}}{4}
Final answer: x=\dfrac{2+\sqrt{10}}{2} or x=\dfrac{2-\sqrt{10}}{2}.
Nature of roots using the discriminant
The discriminant is useful when the question asks whether roots are real, equal, unequal, or not real. You do not need to solve the full equation unless the question asks for the roots.
Worked example: Find the nature of roots of 2x^2-4x+5=0
Step 1: Identify a=2, b=-4, and c=5.
Step 2: Compute D=b^2-4ac.
D=(-4)^2-4(2)(5)
D=16-40=-24
Step 3: Since D<0, the equation has no real roots.
Final answer: The roots are not real.
Word problem model using a quadratic equation
A practical application in Chapter 5 is to form a quadratic equation from a statement. Define the unknown first; then translate the condition into an equation.
Worked example: The product of two consecutive positive integers is 306. Find the integers.
Step 1: Let the smaller positive integer be n. Then the next integer is n+1.
Step 2: Form the equation from the product.
n(n+1)=306
n^2+n-306=0
Step 3: Factorise the quadratic.
n^2+18n-17n-306=0
n(n+18)-17(n+18)=0
(n-17)(n+18)=0
Step 4: Solve and apply the condition that the integers are positive.
n=17 \quad \text{or} \quad n=-18
Final answer: The consecutive positive integers are 17 and 18.
Quick answer index
This table gives a one-line check for the worked questions above. Use it only after reading the steps; in ICSE Class 10 Maths, the working is as important as the final root.
| Section | Question | Answer |
|---|---|---|
| Exercise 5(A) | 1(a) | x=\dfrac{3}{2} or x=-\dfrac{3}{2} |
| Exercise 5(A) | 1(b) | x=3 or x=-5 |
| Exercise 5(A) | 1(c) | k=-3 |
| Exercise 5(A) | 2 | m=-2\sqrt{6} |
| Exercise 5(A) | 3 | m=9,\ n=-15 |
| Exercise 5(B) | 1(a) | x=-1 or x=7 |
| Exercise 5(B) | 2 | x=-2 or x=5 |
| Exercise 5(B) | 3 | x=\sqrt{13} or x=-\sqrt{13} |
| Exercise 5(B) | 4 | x=\dfrac{5}{8} or x=\dfrac{5}{2} |
| Exercise 5(B) | 5 | If x\in\mathbb{N}, x=2; if x\in\mathbb{Q}, x=\dfrac{5}{2} or x=2 |
| Exercise 5(B) | 6 | x=9 or x=-9 |
| Exercise 5(B) | 7 | x=2 or x=-\dfrac{7}{2} |
| Exercise 5(B) | 8 | x=-2 or x=12 |
| Exercise 5(B) | 9 | x=5 or x=-5 |
| Method example | Completing the square | x=3+\sqrt{7} or x=3-\sqrt{7} |
| Method example | Quadratic formula | x=\dfrac{2+\sqrt{10}}{2} or x=\dfrac{2-\sqrt{10}}{2} |
Examiner’s mindset for quadratic equations
In a quadratic equation answer, the examiner looks for a chain of correct algebra, not a guessed pair of roots. A safe answer usually shows these checkpoints: standard form ax^2+bx+c=0, a valid factorisation or formula substitution, both roots, and a final check for restrictions when fractions are present. In word problems, define the variable before forming the equation; otherwise the final number may be unclear even if the algebra is correct.
A common marking loss happens when a student writes x^2=81 and then gives only x=9. Since both 9 and -9 have square 81, the complete answer is x=\pm 9, unless the question gives a condition that rejects one root.
Common mistakes students make in Quadratic Equations
- Forgetting the second root: From x^2=25, write x=\pm 5, not only x=5.
- Ignoring restrictions in fractional equations: In \dfrac{1}{x-1}, the value x=1 is not allowed. State restrictions before cross multiplication.
- Wrong signs in the quadratic formula: If b=-4, then -b=4. Write the substitution carefully.
- Splitting the middle term without checking: For 2x^2-9x+10, the split -4x-5x works because -4-5=-9 and \((-4)(-5)=20=2\times10\).
- Not applying the given number set: If the roots are 2 and \dfrac{5}{2} and the question says x\in\mathbb{N}, only 2 is accepted under the usual positive-integers convention.
Frequently Asked Questions
What is the standard form of a quadratic equation in ICSE Class 10 Maths?
The standard form is ax^2+bx+c=0, where a\ne 0. In ICSE Class 10 Maths, writing the equation in this form helps you identify a, b, and c before using factorisation, completing the square, or the quadratic formula.
When should I use factorisation in Selina Chapter 5 Quadratic Equations?
Use factorisation when the quadratic expression can be split into two simple linear factors. For example, x^2-6x-7=0 becomes \((x+1)(x-7)=0\), so the roots are x=-1 and x=7.
How do I avoid losing a root after taking square roots?
When you reach a step such as x^2=81, write x=\pm 9. A square equation normally gives two possible values, unless the question gives a condition such as positive value only.
What does the discriminant tell in Concise Mathematics Selina Solutions Class 10 ICSE Chapter 5 Quadratic Equations?
The discriminant D=b^2-4ac tells the nature of roots. If D>0, the roots are real and unequal. If D=0, the roots are real and equal. If D<0, the equation has no real roots.
Why must restrictions be written in fractional quadratic equations?
Restrictions are needed because denominators cannot be zero. In an equation containing \dfrac{1}{x-3}, the value x=3 is not allowed. After solving the quadratic, reject any root that violates such a restriction.