Quadratic Equations — Class 10 Maths PYQs

Previous-year board questions (2023–2026) with marking-scheme solutions, grouped by topic and marks.

Try each question first, then press (or tap Show Solution) to reveal the answer. Press again for the next question.

Year:Marks:Level:

Is it quadratic equation or not?

1 Mark Questions
11 Mark · March 2025 · Standardopen ↗
Which of the following equations is a quadratic equation?
  • (a)$x^2+1=(x-1)^2$
  • (b)$(x+\sqrt{x})^2=2x\sqrt{x}$
  • (c)$x^3+3x^2=(x+1)^3$
  • (d)$(x+1)(x-1)=(x+1)^2$
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(B) $(x+\sqrt{x})^2=2x\sqrt{x}$
21 Mark · March 2025 · Basicopen ↗
The equation $x + \frac{1}{x} = 3$ ($x \neq 0$) is expressed as a quadratic equation in the form of $ax^2 + bx + c = 0$. The value of $a - b + c$ is :
  • (a)$5$
  • (b)$2$
  • (c)$1$
  • (d)$- 1$
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(A) $5$
31 Mark · March 2025 · Basicopen ↗
If $x = \sqrt{x}$, ($x \neq 0$) is expressed as a quadratic equation in the form $ax^2 + bx + c = 0$, then the value of $a + b + c$ is :
  • (a)$0$
  • (b)$1$
  • (c)$2$
  • (d)$3$
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(A) $0$
41 Mark · March 2025 · Basicopen ↗
If $(\sqrt{x} + 1)^2 = x^2 + 2\sqrt{x}$ is expressed as a quadratic equation in the form of $ax^2 + bx + c = 0$, then the value of $a - b + c$ is :
  • (a)$-1$
  • (b)$0$
  • (c)$1$
  • (d)$2$
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(C) $1$
51 Mark · March 2025 · Basicopen ↗
If $(x + 1)^3 = x^3 + 1$ is expressed as a quadratic equation in the form $px^2 + qx + r = 0$, then the value of $p - q + r$ is :
  • (a)$0$
  • (b)$1$
  • (c)$3$
  • (d)$6$
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(A) $0$
61 Mark · March 2026 · Basicopen ↗
Which of the equations among the following is/are quadratic equation(s) ?
$q_1 : x^2 + x = (x+1)^2$, $q_2 : x-1= x^2 -1$, $q_3 : x^4 = x^2$, $q_4 : \sqrt{x} = x^2\sqrt{x} +1$
  • (a)$q_1$ only
  • (b)$q_1,q_2$ and $q_3$ only
  • (c)$q_2$ only
  • (d)$q_2$ and $q_4$ only
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(c) $q_2$ only
71 Mark · March 2026 · Basicopen ↗
Which of the following equations is/are quadratic equation(s) ?
$q_1: (y+1)^2 = 2y$,
$q_2: (y-1)^2 = y^2$
$q_3 : (y+1)^3 = (y-1)^3$,
$q_4:1+\sqrt{y} = (\sqrt{y} +1)^2$
  • (a)$q_1, q_2$ and $q_4$
  • (b)$q_1$ and $q_2$
  • (c)$q_1, q_3$ and $q_4$
  • (d)$q_1$ and $q_3$
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(d) $q_1$ and $q_3$
81 Mark · March 2026 · Basicopen ↗
Which of the following equations is/are not quadratic ?
$q_1: (x-1)^2 = x^2$,
$q_2: (x-1)^3 = x^3$
$q_3: (x+1)^3 = 3x^2$,
$q_4: (\sqrt{x}+1)^2 = 2\sqrt{x}$
  • (a)$q_1, q_3$ and $q_4$
  • (b)$q_2$ and $q_3$
  • (c)$q_2, q_3$ and $q_4$
  • (d)$q_1$ and $q_4$
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(a) $q_1, q_3$ and $q_4$

verify the solution of quadratic equation

1 Mark Questions
91 Mark · March 2023 · Standardopen ↗
If $x = 0.3$, is a root of the equation $x^2 -0.9k = 0$, then $k$ is equal to :
  • (a)$1$
  • (b)$10$
  • (c)$0.1$
  • (d)$100$
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(c) $0.1$
101 Mark · March 2023 · Standardopen ↗
A quadratic equation whose roots are $(2 + \sqrt{3})$ and $(2 - \sqrt{3})$ is:
  • (a)$x^2 - 4x + 1 = 0$
  • (b)$x^2 + 4x + 1 = 0$
  • (c)$4x^2-3 = 0$
  • (d)$x^2-1 = 0$
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(a) $x^2 - 4x + 1 = 0$
111 Mark · March 2023 · Standardopen ↗
A quadratic equation whose roots are $(3 - \sqrt{2})$ and $(3 + \sqrt{2})$ is :
  • (a)$x^2 - 6x + 7 = 0$
  • (b)$x^2 + 6x + 7 = 0$
  • (c)$9x^2 - 2 = 0$
  • (d)$x^2 - 7 = 0$
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(a) $x^2 - 6x + 7 = 0$
121 Mark · March 2023 · Standardopen ↗
If the zeroes of the quadratic polynomial $x^2 + (a + 1) x + b$ are $2$ and $-3$, then
  • (a)$a = -7, b = -1$
  • (b)$a = 5, b = -1$
  • (c)$a = 2, b = - 6$
  • (d)$a = 0, b =-6$
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(D) $a = 0, b = - 6$
131 Mark · March 2023 · Standardopen ↗
If 'p' is a root of the quadratic equation $x^2 - (p + q) x + k = 0$, then the value of 'k' is
  • (a)$p$
  • (b)$q$
  • (c)$p+q$
  • (d)$pq$
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(D) $pq$
141 Mark · March 2024 · Standardopen ↗
If $x = 5$ is a solution of the quadratic equation $2x^2 + (k - 1)x + 10 = 0$, then the value of $k$ is:
  • (a)$11$
  • (b)$-11$
  • (c)$13$
  • (d)$-13$
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(B) $-11$
151 Mark · March 2024 · Standardopen ↗
Value of k for which $x = 2$ is a solution of the equation $5x^2 - 4x + (2 + k) = 0$, is
  • (a)$10$
  • (b)$-10$
  • (c)$14$
  • (d)$-14$
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(D) $-14$
161 Mark · March 2024 · Standardopen ↗
If $y = 1$ is one of the solutions of the quadratic equation $py^2 + py + 3 = 0$, then the value of $p$ is :
  • (a)$-3$
  • (b)$2$
  • (c)$-\frac{3}{2}$
  • (d)$-2$
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(C) $-\frac{3}{2}$
171 Mark · March 2025 · Standardopen ↗
The quadratic equation whose roots are $7$ and $\frac{1}{7}$ is :
  • (a)$7x^2-50x + 7 = 0$
  • (b)$7x^2-50x + 1 = 0$
  • (c)$7x^2 + 50x-7=0$
  • (d)$7x^2 + 50x - 1 = 0$
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(A) $7x^2 - 50 x + 7 = 0$
181 Mark · March 2026 · Basicopen ↗
If $x = - 1$ is a root of the equation $ax^2 – bx + 3 = 0$, then :
  • (a)$-a+b-3=0$
  • (b)$a-b-3=0$
  • (c)$-a-b+3=0$
  • (d)$a+b+3 = 0$
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(D) $a + b + 3 = 0$
191 Mark · March 2026 · Basicopen ↗
If $x = \sqrt{3}$ is a solution of the equation $ax^2 + \sqrt{3}x - 12 = 0$, then
  • (a)$a = 3$
  • (b)$a = 2$
  • (c)$a = 1$
  • (d)$a = \sqrt{3}$
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(A) $a = 3$

Word Problems

3 Marks Questions
203 Marks · July 2023 · Standardopen ↗
Two water taps together can fill a tank in $3\frac{1}{3}$ hours. The tap of larger diameter takes $5$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.
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Let the time taken by the tap of smaller diameter to fill the tank separately be '$x$' hours and the time taken by the tap of larger diameter to fill the tank separately be $(x - 5)$ hours.
A.T.Q.
$\frac{1}{x} + \frac{1}{x-5} = \frac{3}{10}$
$\Rightarrow 3x^2 - 35x + 50 = 0$
$\Rightarrow (x - 10) (3x - 5) = 0$
$x = 10$ or $x = \frac{5}{3}$
But $x = \frac{5}{3}$ is not possible, so $x = 10$
$\therefore$ time taken by the tap of smaller diameter to fill the tank separately is $10$ hours
and time taken by the tap of larger diameter to fill the tank separately is $10-5=5$ hours
213 Marks · July 2023 · Standardopen ↗
Sum of the areas of two squares is $468$ m$^2$. If the difference of their perimeters is $24$ m, find the lengths of the sides of the two squares.
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Let the lengths of the sides of the two squares be '$x$' m and '$y$' m s.t. $x > y$
A.T.Q.
$x^2 + y^2 = 468$ -----(1) (1/2 Mark)
$4x - 4y = 24$
$\Rightarrow x-y=6$ -----(2) (1/2 Mark)
From (1) and (2), we get
$y^2 + 6y - 216 = 0$
$\Rightarrow y = 12$ and $y = -18$ (1 Mark)
But side of a square is always positive,
So, $y = 12$
and $x = 18$ (1 Mark)
Hence, the lengths of the sides of two squares are $12$ m and $18$ m.
223 Marks · March 2023 · Standardopen ↗
The sum of two numbers is $15$. If the sum of their reciprocals is $\frac{3}{10}$, find the two numbers.
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Let one number be $x \Rightarrow \text{another number} = 15 - x$
Therefore, $\frac{1}{x} + \frac{1}{15-x} = \frac{3}{10}$
$\frac{15-x + x}{x(15 - x)} = \frac{3}{10} \Rightarrow 150 = 3x(15 - x)$
$3x^2 - 45x + 150 = 0$
$x^2 - 15x + 50 = 0 \Rightarrow (x - 10)(x - 5) = 0$
$\Rightarrow x = 10, 5$
Numbers are $10, 5$ or $5, 10$
233 Marks · March 2024 · Standardopen ↗
In a $2$-digit number, the digit at the unit's place is $5$ less than the digit at the ten's place. The product of the digits is $36$. Find the number.
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Let digit at ten's place be $x$
then digit at unit's place $= x - 5$
$x(x - 5) = 36$
$\Rightarrow x^2 - 5x - 36 = 0$
$(x-9)(x + 4) = 0$
$x \ne -4$ so, $x = 9$
$\therefore$ Required number is $94$
243 Marks · March 2024 · Standardopen ↗
Three consecutive integers are such that sum of the square of second and product of other two is $161$. Find the three integers.
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Let the three numbers be $x, x+1$ and $x+2$ ($\frac{1}{2}$)
$(x + 1)^2 + x(x + 2) = 161$
$\Rightarrow x^2+2x-80 = 0$ (1)
$\Rightarrow (x+10)(x-8)=0$
$\therefore x= 8$ or $-10$ (1)
So, the numbers are $8, 9, 10$ or $-10, -9, -8$ ($\frac{1}{2}$)
253 Marks · March 2024 · Standardopen ↗
A dealer sells an article for ₹75 and gains as much percent as the cost price of the article. Find the cost price of the article.
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Let the cost price of the article be $x$
$\therefore$ Gain % $= x$
$x = \frac{75-x}{x} \times 100$
$\Rightarrow x^2+ 100x - 7500 = 0$
$\Rightarrow (x - 50)(x + 150) = 0$
$x \ne -150 \therefore x = 50$
So, the cost price of the article is ₹50
263 Marks · March 2026 · Standardopen ↗
Find two consecutive negative integers, sum of whose squares is $481$.
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Let two consecutive negative integers be $x$ and $(x + 1)$
According to given statement,
$x^2 + (x + 1)^2 = 481$ (1 Mark)
$\Rightarrow x^2 + x - 240 = 0$ (1/2 Mark)
$\Rightarrow (x + 16) (x - 15) = 0$ (1 Mark)
$x = -16$ or $15$
$x = 15$ does not satisfy the given condition.
So, required integers are $-16$ and $-15$. (1/2 Mark)
273 Marks · March 2025 · Basicopen ↗
The sum of the squares of two consecutive even numbers is $452$. Find the numbers.
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Let the two consecutive even numbers be $x$ and $x + 2$.
A.T.Q.
$x^2 + (x + 2)^2 = 452$
$\Rightarrow x^2 + 2x - 224 = 0$
$\Rightarrow (x + 16)(x - 14) = 0$
$\Rightarrow x = 14$
Required numbers are $14$ and $16$.
283 Marks · March 2025 · Basicopen ↗
A rectangular field is $16$ m long and $10$ m wide. There is a path of equal width all around it, having an area of $120$ sq.m. Find the width of the path.
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Let the width of the path be $x$ m.
A. T. Q. $$\begin{aligned}& (16 + 2x)(10 + 2x) - 16 \times 10 = 120 \\ & \Rightarrow 4x^2 + 52x - 120 = 0 \text{ or } x^2 + 13x - 30 = 0 \\ & \Rightarrow (x - 2)(x + 15) = 0 \\ & \Rightarrow x = 2 \text{ (Rejecting } x = -15) \\ & \therefore \text{Width of the path is } 2 \text{ m.}\end{aligned}$$
293 Marks · March 2025 · Basicopen ↗
The sum of the squares of two consecutive odd numbers is 514. Find the numbers.
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Let two consecutive odd numbers be x and $x + 2$
A.T.Q.
$x^2 + (x + 2)^2 = 514$
$\Rightarrow 2x^2 + 4x - 510 = 0$ or $x^2 + 2x - 255 = 0$
$\Rightarrow (x + 17) (x - 15) = 0$
$\Rightarrow x = 15$
Required numbers are 15 and 17
4 Marks Questions
304 Marks · March 2023 · Standardopen ↗
While designing the school year book, a teacher asked the student that the length and width of a particular photo is increased by $x$ units each to double the area of the photo. The original photo is 18 cm long and 12 cm wide. Based on the above information, answer the following questions:
(I) Write an algebraic equation depicting the above information.
(II) Write the corresponding quadratic equation in standard form.
(III) What should be the new dimensions of the enlarged photo?
OR
Can any rational value of $x$ make the new area equal to $220\text{cm}^2$
figure for this question
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(i) $(18 + x) (12 + x) = 2(18\times12)$
(ii) $x^2 + 30x - 216 = 0$
(iii) Solving : $x^2 + 30x – 216 = 0$
$\Rightarrow (x + 36) (x – 6) = 0$
$x \neq -36 \therefore x = 6$.
new dimensions are $24$ cm $\times 18$ cm
OR
(iii) If $(18+ x) (12 + x) = 220$
then $x^2 + 30x - 4=0$
Here $D = 900 + 16 = 916$ which is not a perfect square.
Thus we can't have any such rational value of $x$.
314 Marks · March 2025 · Standardopen ↗
A garden designer is planning a rectangular lawn that is to be surrounded by a uniform walkway.
The total area of the lawn and the walkway is $360$ square metres. The width of the walkway is same on all sides. The dimensions of the lawn itself are $12$ metres by $10$ metres.
Based on the information given above, answer the following questions:
(i) Formulate the quadratic equation representing the total area of the lawn and the walkway, taking width of walkway $= x \operatorname{m}$.
(ii) (a) Solve the quadratic equation to find the width of the walkway 'x'.
OR
(b) If the cost of paving the walkway at the rate of ₹50 per square metre is ₹12,000, calculate the area of the walkway.
(iii) Find the perimeter of the lawn.
figure for this question
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(i) $(12 + 2x)(10 + 2x) = 360$
$4x^2 + 44x - 240 = 0$ or $x^2 + 11x - 60 = 0$
(ii)(a) $(x + 15)(x - 4) = 0$
$x = 4$
$\therefore$ width of the walkway $= 4 \operatorname{m}$
OR
(ii)(b) Area of the walkway $= \frac{12000}{50}$
$= 240 \operatorname{m}^2$
(iii) Perimeter of the lawn $= 2(12 + 10) = 44 \operatorname{m}$
5 Marks Questions
325 Marks · March 2023 · Standardopen ↗
A train travels at a certain average speed for a distance of $54$ km and then travels a distance of $63$ km at an average speed of $6$ km/h more than the first speed. If it takes $3$ hours to complete the journey, what was its first average speed ?
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Let first average speed of the train be $x$ km/hr.
$\frac{54}{x} + \frac{63}{x + 6} = 3$
$\Rightarrow 54(x + 6) + 63x = 3x^2 + 18x$
$\Rightarrow 3x^2 - 99x - 324 = 0$ or $x^2 - 33x - 108 = 0$
$\Rightarrow (x - 36) (x + 3) = 0$
$\Rightarrow x = 36, -3$ (rejected)
Therefore, first average speed of the train was $36$ km/hr.
335 Marks · March 2023 · Standardopen ↗
Two pipes together can fill a tank in $\frac{15}{8}$ hours. The pipe with larger diameter takes $2$ hours less than the pipe with smaller diameter to fill the tank separately. Find the time in which each pipe can fill the tank separately.
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Let the time taken by smaller diameter tap be $x$ hrs.
Time taken by larger diameter tap is $(x - 2)$ hrs.
Therefore $\frac{1}{x-2} + \frac{1}{x} = \frac{8}{15}$
$\Rightarrow 15(2x - 2) = 8x(x - 2)$
$\Rightarrow 8x^2 - 46x + 30 = 0$
$\Rightarrow 4x^2 - 23x + 15 = 0$
$\Rightarrow (4x - 3)(x - 5) = 0$
$\Rightarrow x = \frac{3}{4}$ as $x-2 < 0$ (rejected) or $x = 5$
Smaller diameter tap fills in $5$ hrs.
Larger diameter tap fills in $3$ hrs.
345 Marks · March 2024 · Standardopen ↗
The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is $\frac{16}{21}$, find the fraction.
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Let numerator be $x$,
then denominator be $(2x + 1)$
Fraction = $\frac{x}{2x+1}$
A.T.Q.
$\frac{x}{2x+1} + \frac{2x+1}{x} = \frac{58}{21}$
$\Rightarrow 11x^2 - 26x - 21 = 0$
$\Rightarrow (x - 3)(11x + 7) = 0$
$x \neq -\frac{7}{11}$ So, $x = 3$
$\therefore$ Fraction = $\frac{3}{7}$
355 Marks · March 2024 · Standardopen ↗
A train travels a distance of $90$ km at a constant speed. Had the speed been $15$ km/h more, it would have taken $30$ minutes less for the journey. Find the original speed of the train.
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Let the original speed be $x$ km/h
New speed $= (x + 15)$ km/h
A.T.Q.
$$\begin{aligned}& \frac{90}{x} - \frac{90}{x+15} = \frac{1}{2} \\ & \Rightarrow x^2 + 15x - 2700 = 0 \\ & \Rightarrow(x + 60) (x - 45) = 0 \\ & x \neq -60 \text{ , } x = 45 \\ & \text{The original speed of the train } = 45\text{km/h}\end{aligned}$$
365 Marks · July 2025 · Standardopen ↗
At present, Sourav's age is $3$ years more than the square of his son Ravi's age. When Ravi grows to his father's present age, Sourav's age would be $6$ years less than $13$ times the present age of Ravi. Find present ages of Ravi and Sourav.
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Let the present age of Ravi be 'r' years
and the present age of Sourav be 's' years
Therefore, $s = 3 + r^2$ --- (1)
Ravi grows to father's present age in $(s - r)$ years.
$\therefore$ father's age after $(s - r)$ years = $(2s - r)$ years
and Ravi's age after $(s - r)$ years = 's' years
Therefore, $2s - r = 13r - 6$ or $s = 7r - 3$ --- (2)
Using (1) and (2),
$r^2 - 7r + 6 = 0$
$\Rightarrow (r - 6)(r - 1) = 0$
$\Rightarrow r = 6, 1$
Ignoring $r = 1$ as $s \neq 4$
r = $6$
Hence $s = 39$
375 Marks · March 2025 · Standardopen ↗
The sum of the areas of two squares is $52$ cm$^2$ and difference of their perimeters is $8$ cm. Find the lengths of the sides of the two squares.
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Let the lengths of the sides of two squares be 'x' cm and 'y' cm such that $x > y$.
ATQ
$x^2 + y^2 = 52$ ----- (1)
$4x - 4y = 8$ or $x - y = 2$ ----- (2)
From (1) and (2), we have
$y^2 + 2y - 24 = 0$
$\Rightarrow (y + 6) (y - 4) = 0$
$\therefore y = 4$
So, $x = 2 + 4 = 6$
$\therefore$ Lengths of the sides of two squares are $6$ cm and $4$ cm respectively.
385 Marks · March 2025 · Standardopen ↗
The numerator of a fraction is $3$ less than its denominator. If $2$ is added to both numerator and denominator, then the sum of the new fraction and the original fraction is $\frac{29}{20}$. Find the original fraction.
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Let denominator be $x$
$\therefore$ Numerator $= (x - 3)$
Therefore, fraction $= \frac{x-3}{x}$ ($1$)
ATQ
$\frac{x-3}{x} + \frac{x-3+2}{x+2} = \frac{29}{20}$ ($1$)
$\Rightarrow 11x^2 - 98x - 120 = 0$ ($1$)
$\Rightarrow (x-10)(11x + 12) = 0$ ($1/2$)
So, $x = 10$ ($1/2$)
$\therefore$ Fraction $= \frac{7}{10}$ ($1$)
395 Marks · March 2025 · Standardopen ↗
A train travelling at a uniform speed for $360$ km would have taken $48$ minutes less to travel the same distance if its speed were $5$ km/h more. Find the original speed of the train.
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Let the original speed of train be 'x' km/h
ATQ
$\frac{360}{x} - \frac{360}{x + 5} = \frac{48}{60}$ ($2$)
$\Rightarrow x^2 + 5x - 2250 = 0$ ($1$)
$\Rightarrow (x + 50)(x - 45) = 0$ ($1$)
So, $x = 45$
$\therefore$ Original speed of the train is $45$ km/h. ($1$)
405 Marks · March 2025 · Standardopen ↗
Find two consecutive odd numbers, sum of whose squares is $650$.
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Let the consecutive odd numbers be $x$ and $x + 2 \implies x^2 + (x + 2)^2 = 650 \implies 2x^2 + 4x - 646 = 0$ or $x^2 + 2x - 323 = 0 \implies (x + 19)(x - 17) = 0 \implies x = 17 \therefore$ Odd numbers are $17, 19$
415 Marks · March 2026 · Standardopen ↗
A faster train takes one hour less than a slower train for a journey of $200$ km. If the speed of the slower train is $10$ km/hr less than that of the faster train, find the speeds of the two trains.
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Let the speed of faster train be $x$ km/h
$\therefore$ speed of slower train $= (x – 10)$ km/h
According to the question,
$\frac{200}{x-10} - \frac{200}{x} = 1$ (I) (2 Marks)
$\Rightarrow x^2 - 10x - 2000 = 0$ (II) (1 Mark)
$\Rightarrow (x - 50)(x + 40) = 0$ (III) (1 Mark)
$\therefore x = 50$
$x = -40$ (Rejected) (IV) (1/2 Mark)
Hence, speed of faster train $= 50$ km/h
and speed of slower train $= 40$ km/h (V) (1/2 Mark)
425 Marks · March 2026 · Standardopen ↗
The sum of the areas of two squares is $640$ m$^2$. If the difference in their perimeters is $64$ m, find the sides of the two squares.
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Let the sides of the two squares be $x$ m and $y$ m $(x > y)$
$x^2 + y^2 = 640$ (I) (1 Mark)
and $4x – 4y = 64 \Rightarrow y = x - 16$ (II) (1 Mark)
$\therefore x^2 + (x – 16)^2 = 640$ (III) (1 Mark)
$\Rightarrow x^2 - 16x – 192 = 0$ (IV) (1 Mark)
$\Rightarrow (x – 24)(x + 8) = 0$
$\therefore x = 24$ (V) (1/2 Mark)
$x = -8$ (Rejected)
$\Rightarrow y = 24 – 16 = 8$ (VI) (1/2 Mark)
Hence the sides of the two squares are $24$ m and $8$ m
435 Marks · March 2026 · Standardopen ↗
Venkat can row a boat in still water at the speed of 12 km/h. He ferries tourists 15 km upstream and 18 km downstream in 3 hours. Find the speed of the stream.
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Let speed of the stream be $x$ km/h
$\frac{15}{12 - x} + \frac{18}{12 + x} = 3$ (I) (2)
$\Rightarrow x^2 - x - 12 = 0$ (II) (1)
$\Rightarrow (x - 4)(x + 3) = 0$ (III) (1)
$\Rightarrow x = 4, -3$
x = -3 (rejected)
$\therefore x = 4$ (IV) (1)
The speed of the stream = 4 km/h
445 Marks · March 2026 · Standardopen ↗
By selling an article for ₹$48$, a trader loses as much percent as half of the cost price of the article. Calculate the cost price and loss amount of the article.
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Let the cost price (CP)of the article be $x$
Loss = $\frac{x}{2}\%$ of $x = \frac{x^2}{200}$ (I Mark)
Selling price = $48 = x - \frac{x^2}{200}$ (II Mark)
$\Rightarrow x^2 - 200x + 9600 = 0$ (III Mark)
$\Rightarrow (x - 120)(x - 80) = 0$ (IV Mark)
$\Rightarrow x = 120,80$ (V Mark)
When CP = ₹$120$, Loss = $120 - 48 = 72$ (VI Mark)
When CP = ₹$80$, Loss = $80 - 48 = 32$ (VII Mark)
455 Marks · March 2026 · Standardopen ↗
Two water taps together can fill a tank in $8\frac{8}{3}$ hours. The tap of larger diameter takes $4$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
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Let tap of smaller diameter takes $x$ hours to fill the tank.
$\therefore$ tap of larger diameter takes $x - 4$ hours to fill the tank.
$\frac{1}{x} + \frac{1}{x-4} = \frac{9}{80}$ (2 Marks)
$\Rightarrow 9x^2-196x + 320 = 0$ (1 Mark)
$\Rightarrow (x-20)(9x – 16) = 0$ (1 Mark)
$\Rightarrow x = 20, \frac{16}{9}$
$x = \frac{16}{9}$ (rejected) (1/2 Mark)
$\therefore x = 20$
Hence, tap of smaller diameter and larger diameter takes $20$ hours & $16$ hours respectively, to fill the tank. (1/2 Mark)

Find roots

1 Mark Questions
461 Mark · March 2024 · Standardopen ↗
The roots of the quadratic equation $x^2 + x - p (p + 1) = 0$ are :
  • (a)$p, p + 1$
  • (b)$-p, p + 1$
  • (c)$-p, -(p + 1)$
  • (d)$p, -(p + 1)$
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(D) $p, -(p + 1)$
471 Mark · March 2026 · Standardopen ↗
The roots of the quadratic equation $(x – 1)^2 = 16$ are :
  • (a)$5,3$
  • (b)$4,-4$
  • (c)$5,-3$
  • (d)$-5,3$
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(C) $5, -3$

Discriminant

4 Marks Questions
484 Marks · March 2024 · Standardopen ↗
A rectangular floor area can be completely tiled with $200$ square tiles. If the side length of each tile is increased by $1$ unit, it would take only $128$ tiles to cover the floor.
(i) Assuming the original length of each side of a tile be $x$ units, make a quadratic equation from the above information.
(ii) Write the corresponding quadratic equation in standard form.
(iii) (a) Find the value of $x$, the length of side of a tile by factorisation.
OR
(b) Solve the quadratic equation for $x$, using quadratic formula.
figure for this question
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(i) $200 x^2 = 128 (x + 1)^2$
(ii) $25x^2 = 16x^2 + 32x + 16$
$\Rightarrow 9x^2 - 32x - 16 = 0$
(iii) (a) $9x^2 - 32x - 16 = 0$
$\Rightarrow (9x + 4) (x - 4) = 0$
$x \neq -\frac{4}{9}$ so, $x = 4$
OR
(iii) (b) $x = \frac{32\pm\sqrt{1024+576}}{18} = \frac{32\pm 40}{18}$
$x \neq -\frac{4}{9}$ so, $x = 4$

Relationship of Roots

1 Mark Questions
491 Mark · July 2024 · Standardopen ↗
If the sum and the product of the roots of the quadratic equation $ax^2 + 6x + 4a = 0$ are equal, then 'a' is equal to :
  • (a)$\frac{3}{2}$
  • (b)$-\frac{3}{2}$
  • (c)$\frac{2}{3}$
  • (d)$-\frac{2}{3}$
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(B) $-\frac{3}{2}$

Nature of Roots

1 Mark Questions
501 Mark · March 2026 · Basicopen ↗
If the quadratic equation $ax^2 + 2bx - c = 0$ ($a \neq 0$) has real and equal roots, then which of the following is true?
  • (a)$b^2-ac$
  • (b)$b^2 = 4ac$
  • (c)$b^2 = ac$
  • (d)$b^2 = -4ac$
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(A) $b^2 = ac$
2 Marks Questions
512 Marks · March 2026 · Standardopen ↗
Verify that roots of the quadratic equation $(p-q)x^2 + (q - r)x + (r - p) = 0$ are equal when $q + r = 2p$.
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Discriminant $(D) = (q - r)^2 - 4 (p - q) (r - p)$ (I) ($\frac{1}{2}$ Mark)
$= (q + r - 2p)^2$ (II) ($\frac{1}{2}$ Mark)
Substituting, $q + r = 2p$
$\Rightarrow D = (2p - 2p)^2 = 0$ (III) ($\frac{1}{2}$ Mark)
$\therefore$ Roots of the given equation are equal. (IV) ($\frac{1}{2}$ Mark)
522 Marks · March 2025 · Basicopen ↗
Find the nature of roots of the equation $3x^2 - 4\sqrt{3}x + 4 = 0$.
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Discriminant $= (-4\sqrt{3})^2 - 4 \times 4 \times 3 = 0$
$\Rightarrow$ The given equation has real and equal roots
3 Marks Questions
533 Marks · March 2025 · Standardopen ↗
Find the value of $p$ for which the quadratic equation $(2p + 1)x^2 - (7p + 2)x + (7p - 3) = 0$ has equal roots. Also, find these roots.
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For equal roots, $D = 0$
$[-(7p + 2)^2] - 4(2p + 1)(7p - 3) = 0 \implies 7p^2 - 24p - 16 = 0 \implies (7p + 4)(p - 4) = 0 \implies p = 4, p = -\frac{4}{7}$
For $p = 4$, the equation is $9x^2 - 30x + 25 = 0$ whose roots are $\frac{5}{3}, \frac{5}{3}$
For $p = -\frac{4}{7}$, the equation is $x^2 - 14x + 49 = 0$ whose roots are $7, 7$
5 Marks Questions
545 Marks · March 2024 · Standardopen ↗
Find the value of 'k' for which the quadratic equation $(k + 1) x^2 - 2 (3k + 1) x + (8k + 1) = 0$ has real and equal roots.
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For real and equal roots.
$[-2(3k + 1)]^2 - 4(k + 1)(8k + 1) = 0$
$\Rightarrow k^2 - 3k = 0$
$\therefore k = 0, k = 3$
555 Marks · March 2024 · Standardopen ↗
Find the value of 'k' for which the quadratic equation $(k + 1)x^2 - 6(k + 1)x + 3(k + 9) = 0, k \neq - 1$ has real and equal roots.
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For real and equal roots, $D = b^2 - 4ac = 0$
$36 (k + 1)^2 - 4 (k + 1)\times 3 (k + 9) = 0$
$\Rightarrow k^2 - 2k - 3 = 0$
$\Rightarrow (k - 3) (k + 1) = 0$
$k \neq - 1$ So, $k = 3$
565 Marks · March 2025 · Standardopen ↗
Find the smallest value of $p$ for which the quadratic equation $x^2 - 2(p+1)x + p^2 = 0$ has real roots. Hence, find the roots of the equation so obtained.
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For real roots, $D \geq 0$. $[-2(p+1)]^2 - 4p^2 \geq 0 \Rightarrow p \geq -\frac{1}{2}$ ($\frac{1}{2} + \frac{1}{2} + 1$ marks). $\therefore$ smallest value of $p = -\frac{1}{2}$ ($\frac{1}{2}$ mark). At $p = -\frac{1}{2}$ given equation becomes $x^2 - 2(-\frac{1}{2} + 1)x + (-\frac{1}{2})^2 = 0$ ($\frac{1}{2}$ mark). $x^2 - x + \frac{1}{4} = 0$ or $4x^2 - 4x + 1 = 0$ (1 mark). $(2x-1)(2x-1) = 0$. $\therefore$ roots are $\frac{1}{2}, \frac{1}{2}$ ($\frac{1}{2} + \frac{1}{2}$ marks).
575 Marks · March 2026 · Basicopen ↗
Express $x - \frac{1}{x} = 3$ as a quadratic equation in standard form and hence find its roots. Also, find the value of 'a' for which the equation $x + \frac{1}{x} = a$, when expressed as a quadratic equation, has real and equal roots.
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$x - \frac{1}{x} = 3 \implies x^2 - 1 = 3x \implies x^2 - 3x - 1 = 0$ (1 Mark)
$D = (-3)^2 - 4(1)(-1) = 9 + 4 = 13$ (1 Mark)
Roots are $\frac{3 + \sqrt{13}}{2}$, $\frac{3 - \sqrt{13}}{2}$ (1 Mark)
Now, $x + \frac{1}{x} = a \implies x^2 - ax + 1 = 0$ (1/2 Mark)
Since roots are real and equal, $D = 0$
$(-a)^2 - 4(1)(1) = 0 \implies a^2 - 4 = 0$ (1/2 Mark)
$a^2 = 4 \implies a = 2$ or $a = -2$ (1/2+1/2 Mark)