CBSE Class 10 · Maths
CBSE Class 10 Maths Quadratic Equations Previous Year Questions
Help your child master CBSE Class 10 Maths Quadratic Equations Previous Year Questions with this carefully curated collection drawn from real board papers spanning 2019–2025. Every question comes with a detailed step-by-step solution — covering the discriminant, nature of roots, Vieta’s formulas, and word problems — so your child is fully prepared for the board exam.
CBSE Class 10 Maths Quadratic Equations — Questions with Solutions
The roots of the equation $$ x^2 + 3x – 10 = 0 $$ are:
Solution
Ans. Option (A) is correct.
Explanation: We have,
$$x^2 + 3x – 10 = 0$$ Factorizing:
$$x^2 + 5x – 2x – 10 = 0$$ $$x(x + 5) – 2(x + 5) = 0$$ $$(x – 2)(x + 5) = 0$$ Thus, $x = 2,\ -5$.
Explanation: We have,
$$x^2 + 3x – 10 = 0$$ Factorizing:
$$x^2 + 5x – 2x – 10 = 0$$ $$x(x + 5) – 2(x + 5) = 0$$ $$(x – 2)(x + 5) = 0$$ Thus, $x = 2,\ -5$.
If the quadratic equation $$ ax^2 + bx + c = 0 $$ has two real and equal roots, then ‘$c$’ is equal to:
Solution
Ans. Option (D) is correct.
Explanation: For an equation having real and equal roots, the discriminant $D$ is:
$$D = b^2 – 4ac = 0$$ Thus, $b^2 = 4ac$, and therefore: $$\frac{b^2}{4a} = c$$
Explanation: For an equation having real and equal roots, the discriminant $D$ is:
$$D = b^2 – 4ac = 0$$ Thus, $b^2 = 4ac$, and therefore: $$\frac{b^2}{4a} = c$$
What is/are the roots of $$ 3x^2 = 6x $$
Solution
Ans. Option (D) is correct.
Explanation: We have, $3x^2 = 6x$.
Rearranging: $3x^2 – 6x = 0$
Factoring: $3x(x – 2) = 0$
Thus, $x = 0$ or $x = 2$. So, the roots are 0 and 2.
Explanation: We have, $3x^2 = 6x$.
Rearranging: $3x^2 – 6x = 0$
Factoring: $3x(x – 2) = 0$
Thus, $x = 0$ or $x = 2$. So, the roots are 0 and 2.
If the quadratic equation $$ x^2 – 8x + k = 0 $$ has real roots, then:
Solution
Ans. Option (B) is correct.
Explanation: Comparing with $ax^2 + bx + c = 0$, we get $a = 1,\ b = -8,\ c = k$.
For real roots, $D \geq 0$:
$$(-8)^2 – 4(1)(k) \geq 0$$ $$64 – 4k \geq 0$$ $$k \leq 16$$
Explanation: Comparing with $ax^2 + bx + c = 0$, we get $a = 1,\ b = -8,\ c = k$.
For real roots, $D \geq 0$:
$$(-8)^2 – 4(1)(k) \geq 0$$ $$64 – 4k \geq 0$$ $$k \leq 16$$
The nature of roots of the quadratic equation $$ 9x^2 – 6x – 2 = 0 $$ is:
Solution
Ans. Option (C) is correct.
Explanation: Here, $a = 9,\ b = -6,\ c = -2$.
$$D = b^2 – 4ac = (-6)^2 – 4 \times 9 \times (-2) = 36 + 72 = 108$$ Since $D > 0$, the equation has 2 distinct real roots.
Explanation: Here, $a = 9,\ b = -6,\ c = -2$.
$$D = b^2 – 4ac = (-6)^2 – 4 \times 9 \times (-2) = 36 + 72 = 108$$ Since $D > 0$, the equation has 2 distinct real roots.
Let $p$ be a prime number. The quadratic equation having its roots as factors of $p$ is:
Solution
Ans. Option (B) is correct.
Explanation: The factors of a prime $p$ are $p \times 1$. Thus, the roots are $p$ and $1$.
The quadratic equation is: $$x^2 – (\text{sum of roots})x + \text{product of roots} = 0$$ $$x^2 – (p + 1)x + p = 0$$
Explanation: The factors of a prime $p$ are $p \times 1$. Thus, the roots are $p$ and $1$.
The quadratic equation is: $$x^2 – (\text{sum of roots})x + \text{product of roots} = 0$$ $$x^2 – (p + 1)x + p = 0$$
The value(s) of $k$ for which the quadratic equation $$ 2x^2 + kx + 2 = 0 $$ has equal roots, is:
Solution
Sol. Option (B) is correct.
Explanation: Comparing $2x^2 + kx + 2 = 0$ with $ax^2 + bx + c = 0$, we get $a = 2,\ b = k,\ c = 2$.
For equal roots, $D = 0$:
$$k^2 – 4 \times 2 \times 2 = 0$$ $$k^2 – 16 = 0 \Rightarrow (k + 4)(k – 4) = 0$$ Therefore, $k = \pm 4$.
Explanation: Comparing $2x^2 + kx + 2 = 0$ with $ax^2 + bx + c = 0$, we get $a = 2,\ b = k,\ c = 2$.
For equal roots, $D = 0$:
$$k^2 – 4 \times 2 \times 2 = 0$$ $$k^2 – 16 = 0 \Rightarrow (k + 4)(k – 4) = 0$$ Therefore, $k = \pm 4$.
The discriminant of the quadratic equation $x^2 + 3x – 2 = 0$ is:
Solution
Answer: (B) 17
Explanation: The discriminant $\Delta$ of $ax^2 + bx + c = 0$ is $\Delta = b^2 – 4ac$.
For $x^2 + 3x – 2 = 0$, we have $a = 1,\ b = 3,\ c = -2$: $$\Delta = 3^2 – 4(1)(-2) = 9 + 8 = 17$$
Explanation: The discriminant $\Delta$ of $ax^2 + bx + c = 0$ is $\Delta = b^2 – 4ac$.
For $x^2 + 3x – 2 = 0$, we have $a = 1,\ b = 3,\ c = -2$: $$\Delta = 3^2 – 4(1)(-2) = 9 + 8 = 17$$
The pair of linear equations 2x = 5y + 6 and 15y = 6x − 18 represents two lines which are:
Solution
Ans. Option (C) is correct.
Explanation: From the given equations: \[\frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-5}{-15} = \frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{-6}{-18} = \frac{1}{3}\] Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, the lines are coincident.
Explanation: From the given equations: \[\frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-5}{-15} = \frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{-6}{-18} = \frac{1}{3}\] Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, the lines are coincident.
3 chairs and 1 table cost ₹900; whereas 5 chairs and 3 tables cost ₹2,100. If the cost of 1 chair is ₹$x$ and the cost of 1 table is ₹$y$, then the situation can be represented algebraically as:
Solution
Ans. Option (C) is correct.
Explanation: Given cost of one chair = $x$ and cost of one table = $y$.
• 3 chairs + 1 table = ₹900 → $3x + y = 900$
• 5 chairs + 3 tables = ₹2100 → $5x + 3y = 2100$
Explanation: Given cost of one chair = $x$ and cost of one table = $y$.
• 3 chairs + 1 table = ₹900 → $3x + y = 900$
• 5 chairs + 3 tables = ₹2100 → $5x + 3y = 2100$
If the roots of the equation $ax^2 + bx + c = 0$, $a \neq 0$ are real and equal, then which of the following relations is true?
Solution
Answer: Option (C) is correct.
Explanation: For real and equal roots, discriminant $= 0$: $$b^2 – 4ac = 0 \Rightarrow b^2 = 4ac \Rightarrow ac = \frac{b^2}{4}$$
Explanation: For real and equal roots, discriminant $= 0$: $$b^2 – 4ac = 0 \Rightarrow b^2 = 4ac \Rightarrow ac = \frac{b^2}{4}$$
The discriminant of the quadratic equation $3\sqrt{3}\,x^2 + 10x + \sqrt{3} = 0$ is:
Solution
Answer: Option (D) is correct.
Explanation: For $3\sqrt{3}\,x^2 + 10x + \sqrt{3} = 0$, we have $a = 3\sqrt{3},\ b = 10,\ c = \sqrt{3}$: $$\Delta = b^2 – 4ac = (10)^2 – 4(3\sqrt{3})(\sqrt{3}) = 100 – 4 \times 9 = 100 – 36 = 64$$
Explanation: For $3\sqrt{3}\,x^2 + 10x + \sqrt{3} = 0$, we have $a = 3\sqrt{3},\ b = 10,\ c = \sqrt{3}$: $$\Delta = b^2 – 4ac = (10)^2 – 4(3\sqrt{3})(\sqrt{3}) = 100 – 4 \times 9 = 100 – 36 = 64$$
The least positive value of $k$, for which the quadratic equation $2x^2 + kx – 4 = 0$ has rational roots, is:
Solution
Answer: Option (B) is correct.
Explanation: For rational roots, $\Delta = k^2 – 4(2)(-4) = k^2 + 32$ must be a perfect square.
Let $k^2 + 32 = m^2$, so $(m – k)(m + k) = 32$. The factor pair $(4, 8)$ gives $m – k = 4$ and $m + k = 8$, which yields $k = 2$.
The least positive value of $k$ is 2.
Explanation: For rational roots, $\Delta = k^2 – 4(2)(-4) = k^2 + 32$ must be a perfect square.
Let $k^2 + 32 = m^2$, so $(m – k)(m + k) = 32$. The factor pair $(4, 8)$ gives $m – k = 4$ and $m + k = 8$, which yields $k = 2$.
The least positive value of $k$ is 2.
The value(s) of $k$ for which the quadratic equation $2x^2 + kx + 2 = 0$ has equal roots, is:
Solution
Answer: Option (C) is correct.
Explanation: For $2x^2 + kx + 2 = 0$, we have $a = 2,\ b = k,\ c = 2$.
For equal roots, $D = 0$: $$k^2 – 4(2)(2) = 0 \Rightarrow k^2 = 16 \Rightarrow k = \pm 4$$ Both $k = 4$ and $k = -4$ are valid; the option listed here is $k = -4$, corresponding to Option (C).
Explanation: For $2x^2 + kx + 2 = 0$, we have $a = 2,\ b = k,\ c = 2$.
For equal roots, $D = 0$: $$k^2 – 4(2)(2) = 0 \Rightarrow k^2 = 16 \Rightarrow k = \pm 4$$ Both $k = 4$ and $k = -4$ are valid; the option listed here is $k = -4$, corresponding to Option (C).
If one root of the equation $$(k – 1)x^2 – 10x + 3 = 0$$ is the reciprocal of the other, then find the value of $k$.
Answer
Let one root $= \alpha$ and the other root $= \dfrac{1}{\alpha}$.
Then, product of roots $= \alpha \cdot \dfrac{1}{\alpha} = 1$.
Comparing $(k-1)x^2 – 10x + 3 = 0$ with $ax^2 + bx + c = 0$, we get $a = k-1,\ c = 3$.
Product of roots $= \dfrac{c}{a} = \dfrac{3}{k-1}$
$$\frac{3}{k-1} = 1 \Rightarrow 3 = k – 1 \Rightarrow k = 4$$ Answer: $k = 4$
Then, product of roots $= \alpha \cdot \dfrac{1}{\alpha} = 1$.
Comparing $(k-1)x^2 – 10x + 3 = 0$ with $ax^2 + bx + c = 0$, we get $a = k-1,\ c = 3$.
Product of roots $= \dfrac{c}{a} = \dfrac{3}{k-1}$
$$\frac{3}{k-1} = 1 \Rightarrow 3 = k – 1 \Rightarrow k = 4$$ Answer: $k = 4$
Write the discriminant of the quadratic equation $$(x + 5)^2 = 2(8x – 3).$$
Answer
Expanding both sides:
$$x^2 + 10x + 25 = 16x – 6$$ Rearranging to standard form:
$$x^2 – 6x + 31 = 0$$ Here $a = 1,\ b = -6,\ c = 31$. The discriminant is:
$$\Delta = (-6)^2 – 4(1)(31) = 36 – 124 = \boxed{-88}$$
$$x^2 + 10x + 25 = 16x – 6$$ Rearranging to standard form:
$$x^2 – 6x + 31 = 0$$ Here $a = 1,\ b = -6,\ c = 31$. The discriminant is:
$$\Delta = (-6)^2 – 4(1)(31) = 36 – 124 = \boxed{-88}$$
For what values of $k$, the given quadratic equation $$ 3x^2 + 6kx + 4 = 0 $$ has equal roots?
Answer
For equal roots, discriminant $D = 0$.
Here $a = 3,\ b = 6k,\ c = 4$: $$D = (6k)^2 – 4(3)(4) = 36k^2 – 48 = 0$$ $$36k^2 = 48 \Rightarrow k^2 = \frac{4}{3} \Rightarrow k = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}$$ Answer: $k = \pm\dfrac{2\sqrt{3}}{3}$
Here $a = 3,\ b = 6k,\ c = 4$: $$D = (6k)^2 – 4(3)(4) = 36k^2 – 48 = 0$$ $$36k^2 = 48 \Rightarrow k^2 = \frac{4}{3} \Rightarrow k = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}$$ Answer: $k = \pm\dfrac{2\sqrt{3}}{3}$
For what value(s) of $a$, the quadratic equation $$ 3a^2 – 6a + 1 = 0 $$ has no real roots?
Answer
For no real roots, discriminant $D < 0$:
$$D = b^2 – 4ac = (-6)^2 – 4(3)(1) = 36 – 12a < 0$$
$$12a > 36 \Rightarrow a > 3$$ Answer: The equation has no real roots when $a > 3$.
Find the sum and product of the roots of the quadratic equation $$ 2x^2 – 9x + 4 = 0. $$
Answer
Factorising:
$$2x^2 – 8x – x + 4 = 2x(x-4) – 1(x-4) = (2x-1)(x-4) = 0$$ So the roots are $x = \dfrac{1}{2}$ and $x = 4$.
• Sum $= \dfrac{1}{2} + 4 = \dfrac{9}{2}$
• Product $= \dfrac{1}{2} \times 4 = 2$
$$2x^2 – 8x – x + 4 = 2x(x-4) – 1(x-4) = (2x-1)(x-4) = 0$$ So the roots are $x = \dfrac{1}{2}$ and $x = 4$.
• Sum $= \dfrac{1}{2} + 4 = \dfrac{9}{2}$
• Product $= \dfrac{1}{2} \times 4 = 2$
Find the discriminant of the quadratic equation $4x^2 – 5 = 0$ and hence comment on the nature of roots of the equation.
Answer
Comparing $4x^2 – 5 = 0$ with $ax^2 + bx + c = 0$, we get $a = 4,\ b = 0,\ c = -5$.
$$D = b^2 – 4ac = 0 – 4(4)(-5) = 80 > 0$$ Since $D > 0$, the equation has real and distinct roots.
$$D = b^2 – 4ac = 0 – 4(4)(-5) = 80 > 0$$ Since $D > 0$, the equation has real and distinct roots.
If the sum of the roots of the quadratic equation $ky^2 – 11y + (k – 23) = 0$ is 15 more than the product of the roots, then find the value of $k$.
Answer
Sum of roots $= \dfrac{11}{k}$ and Product of roots $= \dfrac{k-23}{k}$.
Given: Sum $=$ Product $+ 15$: $$\frac{11}{k} = \frac{k-23}{k} + 15$$ $$11 = k – 23 + 15k \Rightarrow 11 = 16k – 23 \Rightarrow 16k = 34 \Rightarrow k = \frac{17}{8}$$ Answer: $k = \dfrac{17}{8}$
Given: Sum $=$ Product $+ 15$: $$\frac{11}{k} = \frac{k-23}{k} + 15$$ $$11 = k – 23 + 15k \Rightarrow 11 = 16k – 23 \Rightarrow 16k = 34 \Rightarrow k = \frac{17}{8}$$ Answer: $k = \dfrac{17}{8}$
If $x = -2$ is the common solution of quadratic equations $ax^2 + x – 3a = 0$ and $x^2 + bx + b = 0$, then find the value of $a \times b$.
Answer
Substituting $x = -2$ in $ax^2 + x – 3a = 0$:
$$4a – 2 – 3a = 0 \Rightarrow a = 2$$ Substituting $x = -2$ in $x^2 + bx + b = 0$:
$$4 – 2b + b = 0 \Rightarrow b = 4$$ $$a \times b = 2 \times 4 = \boxed{8}$$
$$4a – 2 – 3a = 0 \Rightarrow a = 2$$ Substituting $x = -2$ in $x^2 + bx + b = 0$:
$$4 – 2b + b = 0 \Rightarrow b = 4$$ $$a \times b = 2 \times 4 = \boxed{8}$$
Find the value of $k$ for which the quadratic equation $2kx^2 – 40x + 25 = 0$ has real and equal roots.
Answer
Comparing with $ax^2 + bx + c = 0$: $a = 2k,\ b = -40,\ c = 25$.
For real and equal roots, $D = 0$: $$(-40)^2 – 4(2k)(25) = 0$$ $$1600 – 200k = 0 \Rightarrow k = \frac{1600}{200} = \boxed{8}$$
For real and equal roots, $D = 0$: $$(-40)^2 – 4(2k)(25) = 0$$ $$1600 – 200k = 0 \Rightarrow k = \frac{1600}{200} = \boxed{8}$$
Solve for $x$: $$\frac{5}{2}x^2 + \frac{2}{5} = 1 – 2x$$
Answer
Multiplying through by 10 to clear fractions:
$$25x^2 + 4 = 10(1 – 2x) \Rightarrow 25x^2 + 20x – 6 = 0$$
Using the quadratic formula with $a = 25,\ b = 20,\ c = -6$:
$$x = \frac{-20 \pm \sqrt{400 + 600}}{50} = \frac{-20 \pm 10\sqrt{10}}{50} = \frac{-2 \pm \sqrt{10}}{5}$$ Answer: $x = \dfrac{-2 + \sqrt{10}}{5}$ or $x = \dfrac{-2 – \sqrt{10}}{5}$
Find the value of ‘$p$’ for which the quadratic equation $px(x – 2) + 6 = 0$ has two equal real roots.
Answer
Expanding: $px^2 – 2px + 6 = 0$.
For real and equal roots, $D = 0$: $$(-2p)^2 – 4(p)(6) = 0 \Rightarrow 4p^2 – 24p = 0 \Rightarrow 4p(p – 6) = 0$$ So $p = 0$ or $p = 6$. Since the equation must be quadratic, $p = 0$ is rejected.
Answer: $p = 6$
For real and equal roots, $D = 0$: $$(-2p)^2 – 4(p)(6) = 0 \Rightarrow 4p^2 – 24p = 0 \Rightarrow 4p(p – 6) = 0$$ So $p = 0$ or $p = 6$. Since the equation must be quadratic, $p = 0$ is rejected.
Answer: $p = 6$
The sum of two numbers is 15. If the sum of their reciprocals is $\dfrac{3}{10}$, find the two numbers.
Answer
Let the first number be $x$. Then the second number is $15 – x$.
$$\frac{1}{x} + \frac{1}{15 – x} = \frac{3}{10} \Rightarrow \frac{15}{x(15-x)} = \frac{3}{10}$$ $$150 = 3x(15-x) \Rightarrow x^2 – 15x + 50 = 0$$ $$(x – 10)(x – 5) = 0 \Rightarrow x = 10 \text{ or } x = 5$$ Answer: The two numbers are 10 and 5.
$$\frac{1}{x} + \frac{1}{15 – x} = \frac{3}{10} \Rightarrow \frac{15}{x(15-x)} = \frac{3}{10}$$ $$150 = 3x(15-x) \Rightarrow x^2 – 15x + 50 = 0$$ $$(x – 10)(x – 5) = 0 \Rightarrow x = 10 \text{ or } x = 5$$ Answer: The two numbers are 10 and 5.
If $a$ and $b$ are roots of the quadratic equation $x^2 – 7x + 10 = 0$, find the quadratic equation whose roots are $a^2$ and $b^2$.
Answer
Factorising $x^2 – 7x + 10 = 0$:
$$(x – 5)(x – 2) = 0 \Rightarrow a = 5,\ b = 2$$
So $a^2 = 25$ and $b^2 = 4$.
The new quadratic equation with roots $a^2$ and $b^2$: $$x^2 – (a^2 + b^2)x + a^2 b^2 = 0$$ $$x^2 – 29x + 100 = 0$$ Answer: $x^2 – 29x + 100 = 0$
The new quadratic equation with roots $a^2$ and $b^2$: $$x^2 – (a^2 + b^2)x + a^2 b^2 = 0$$ $$x^2 – 29x + 100 = 0$$ Answer: $x^2 – 29x + 100 = 0$
A train covers a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the original speed of the train.
Answer
Let the original speed $= x$ km/h.
Time at original speed $= \dfrac{480}{x}$; time at reduced speed $= \dfrac{480}{x-8}$.
$$\frac{480}{x-8} – \frac{480}{x} = 3 \Rightarrow \frac{3840}{x(x-8)} = 3$$ $$3840 = 3x(x-8) \Rightarrow x^2 – 8x – 1280 = 0$$ Using the quadratic formula: $$x = \frac{8 \pm \sqrt{64 + 5120}}{2} = \frac{8 \pm \sqrt{5184}}{2} = \frac{8 \pm 72}{2}$$ Taking the positive value: $x = \dfrac{8 + 72}{2} = 40$.
Answer: The original speed of the train is $\boxed{40}$ km/h.
Time at original speed $= \dfrac{480}{x}$; time at reduced speed $= \dfrac{480}{x-8}$.
$$\frac{480}{x-8} – \frac{480}{x} = 3 \Rightarrow \frac{3840}{x(x-8)} = 3$$ $$3840 = 3x(x-8) \Rightarrow x^2 – 8x – 1280 = 0$$ Using the quadratic formula: $$x = \frac{8 \pm \sqrt{64 + 5120}}{2} = \frac{8 \pm \sqrt{5184}}{2} = \frac{8 \pm 72}{2}$$ Taking the positive value: $x = \dfrac{8 + 72}{2} = 40$.
Answer: The original speed of the train is $\boxed{40}$ km/h.
Solve for $x$: $$\frac{1}{x + 4} – \frac{1}{x + 7} = \frac{11}{30}, \quad x \neq -4,\ x \neq -7.$$
Answer
$$\frac{(x+7) – (x+4)}{(x+4)(x+7)} = \frac{11}{30} \Rightarrow \frac{3}{x^2 + 11x + 28} = \frac{11}{30}$$
$$11x^2 + 121x + 308 = 90 \Rightarrow 11x^2 + 121x + 218 = 0$$
Using the quadratic formula with $a = 11,\ b = 121,\ c = 218$:
$$x = \frac{-121 \pm \sqrt{14641 – 9592}}{22} = \frac{-121 \pm \sqrt{5049}}{22} \approx \frac{-121 \pm 71.06}{22}$$
$$x \approx -2.27 \quad \text{or} \quad x \approx -8.73$$ Answer: $x \approx -2.27$ or $x \approx -8.73$
Find the value of ‘$p$’ for which the quadratic equation $$p(x – 4)(x – 2) + (x – 1)^2 = 0$$ has real and equal roots.
Answer
Expanding:
$$p(x^2 – 6x + 8) + (x^2 – 2x + 1) = 0$$
$$(p+1)x^2 – 2(3p+1)x + (8p+1) = 0$$
For real and equal roots, $D = 0$:
$$[-2(3p+1)]^2 – 4(p+1)(8p+1) = 0$$
$$4(9p^2 + 6p + 1) – 4(8p^2 + 9p + 1) = 0$$
$$36p^2 + 24p + 4 – 32p^2 – 36p – 4 = 0$$
$$4p^2 – 12p = 0 \Rightarrow 4p(p – 3) = 0$$
$$p = 0 \quad \text{or} \quad p = 3$$ Answer: For $p = 0$ or $p = 3$, the equation has real and equal roots.
The sum of two numbers is 34. If 3 is subtracted from one number and 2 is added to another, the product of these two numbers becomes 260. Find the numbers.
Answer
Let the first number be $x$ and the second be $y$.
$x + y = 34 \Rightarrow y = 34 – x$ and $(x-3)(y+2) = 260$.
Substituting: $$(x-3)(36-x) = 260$$ $$36x – x^2 – 108 + 3x = 260 \Rightarrow x^2 – 39x + 368 = 0$$ $$x = \frac{39 \pm \sqrt{1521 – 1472}}{2} = \frac{39 \pm 7}{2}$$ So $x = 23$ or $x = 16$.
When $x = 23$, $y = 11$. When $x = 16$, $y = 18$.
Answer: The two numbers are 23 and 11, or 16 and 18.
$x + y = 34 \Rightarrow y = 34 – x$ and $(x-3)(y+2) = 260$.
Substituting: $$(x-3)(36-x) = 260$$ $$36x – x^2 – 108 + 3x = 260 \Rightarrow x^2 – 39x + 368 = 0$$ $$x = \frac{39 \pm \sqrt{1521 – 1472}}{2} = \frac{39 \pm 7}{2}$$ So $x = 23$ or $x = 16$.
When $x = 23$, $y = 11$. When $x = 16$, $y = 18$.
Answer: The two numbers are 23 and 11, or 16 and 18.
Frequently Asked Questions
What does the Quadratic Equations chapter cover in CBSE Class 10 Maths? ⌄
The Quadratic Equations chapter covers the standard form $ax^2 + bx + c = 0$, methods of solving quadratic equations (factorisation, completing the square, and the quadratic formula), the discriminant and its role in determining the nature of roots, and real-life word problems. It builds directly on linear equations and is foundational for higher algebra.
How many marks does the Quadratic Equations chapter carry in the CBSE Class 10 board exam? ⌄
Quadratic Equations is part of the Algebra unit, which carries approximately 20 marks in the CBSE Class 10 Mathematics board paper. Questions from this chapter appear across all question types — 1-mark MCQs, 2-mark short answers, 3-mark problems, and 5-mark word problems — making it one of the highest-weightage chapters for your child to prepare thoroughly.
What are the most important topics in Class 10 Quadratic Equations for board exams? ⌄
The most frequently tested topics are: finding the discriminant and determining the nature of roots, finding the value of an unknown constant (like $k$ or $p$) when equal or real roots are given, solving quadratic equations using factorisation and the quadratic formula, and applying Vieta’s formulas for sum and product of roots. Word problems involving speed, distance, and number-based scenarios also appear regularly in 5-mark questions.
What are the most common mistakes students make in Quadratic Equations? ⌄
A very common error is incorrectly identifying $a$, $b$, and $c$ when the equation is not yet in standard form — students must always rearrange first. Another frequent mistake is applying the equal roots condition ($D = 0$) when the question asks for real roots ($D \geq 0$), and vice versa. In word problems, students often set up the equation correctly but make sign errors during simplification, losing marks in the working steps.
How does Angle Belearn help students score well in Quadratic Equations? ⌄
Angle Belearn’s CBSE specialists curate chapter-wise question banks drawn directly from real board papers, each paired with clear, step-by-step solutions that mirror the structured working examiners expect. Regular practice with these verified questions builds your child’s speed, accuracy, and confidence — so they walk into the board exam knowing exactly how to approach every type of Quadratic Equations question.
