---
title: "Quadratic Equations: CBSE Class 10 Maths previous year questions"
url: https://www.swavid.com/cbse/class-10/maths/pyq/quadratic-equations
---

# Quadratic Equations: CBSE Class 10 Maths previous year questions

25 questions from CBSE Class 10 board papers, newest first, each with full working.

## CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions

### Question 33

*5 marks · Long answer*

(A) A faster train takes one hour less than a slower train for a journey of $200\text{ km}$. If the speed of the slower train is $10\text{ km/hr}$ less than that of the faster train, find the speeds of the two trains.

**Solution**

1. Let the speed of the slower train be $x\text{ km/hr}$.
2. Then the speed of the faster train is $(x + 10)\text{ km/hr}$.
3. Distance to be covered = $200\text{ km}$.
4. Time taken by the slower train = $\frac{200}{x}\text{ hours}$ and time taken by the faster train = $\frac{200}{x + 10}\text{ hours}$.
5. According to the given condition, $\frac{200}{x} - \frac{200}{x + 10} = 1$.
6. Simplifying the equation gives $200(x + 10) - 200x = x(x + 10)$, which leads to $x^2 + 10x - 2000 = 0$.
7. Solving by splitting the middle term: $x^2 + 50x - 40x - 2000 = 0$, so $(x + 50)(x - 40) = 0$.
8. Since speed cannot be negative, $x = 40$.
9. Thus, the speed of the slower train is $40\text{ km/hr}$ and the speed of the faster train is $40 + 10 = 50\text{ km/hr}$.

**Answer:** Speed of slower train = $40\text{ km/hr}$, Speed of faster train = $50\text{ km/hr}$.

> Common mistake: Taking speed of slower train as $x - 10$ and faster as $x$ and getting negative time differences.

### Question 33 (OR)

*5 marks · Long answer*

(B) The sum of the areas of two squares is $640\text{ m}^2$. If the difference in their perimeters is $64\text{ m}$, find the sides of the two squares.

**Solution**

1. Let the side of the first square be $x\text{ m}$ and the side of the second square be $y\text{ m}$, where $x > y$.
2. Perimeter of the first square is $4x$ and of the second is $4y$.
3. Given the difference in perimeters is $64\text{ m}$, so $4x - 4y = 64$, which gives $x - y = 16$ or $x = y + 16$.
4. Given the sum of their areas is $640\text{ m}^2$, so $x^2 + y^2 = 640$.
5. Substitute $x = y + 16$ into the area equation: $(y + 16)^2 + y^2 = 640$.
6. Expanding and simplifying: $y^2 + 32y + 256 + y^2 = 640$, which gives $2y^2 + 32y - 384 = 0$.
7. Divide by 2: $y^2 + 16y - 192 = 0$.
8. Factorising the quadratic equation: $y^2 + 24y - 8y - 192 = 0$, so $(y + 24)(y - 8) = 0$.
9. Since side length cannot be negative, $y = 8$.
10. Then $x = 8 + 16 = 24$.
11. The sides of the two squares are $24\text{ m}$ and $8\text{ m}$.

**Answer:** Sides of the two squares are $24\text{ m}$ and $8\text{ m}$.

> Common mistake: Forgetting to divide the perimeter equation by 4 or expanding $(y+16)^2$ incorrectly.

## CBSE Class 10 Maths Basic Question Paper 2025 (Set 430/1/1) with Solutions

### Question 3

*1 mark · MCQ*

The discriminant of the quadratic equation $x^{2}-3x-2=0$ is:

- 1
- 17
- $\sqrt{17}$
- $-\sqrt{17}$

**Solution**

1. Compare the given equation $x^{2}-3x-2=0$ with the standard quadratic equation $ax^{2}+bx+c=0$ to get $a=1$, $b=-3$, and $c=-2$.
2. The discriminant $D$ is given by $b^{2}-4ac = (-3)^{2}-4(1)(-2) = 9 + 8 = 17$.

**Answer:** (b) 17

> Common mistake: Sign errors while squaring negative coefficients or multiplying by $-4ac$.

### Question 4

*1 mark · MCQ*

The equation $x+\frac{1}{x}=3(x\ne0)$ is expressed as a quadratic equation in the form of $ax^{2}+bx+c=0$. The value of $a-b+c$ is:

- 5
- 2
- 1
- -1

**Solution**

1. Given the equation $x+\frac{1}{x}=3$, multiply both sides by $x$ to get $x^{2}+1=3x$.
2. Rearrange into standard form $ax^{2}+bx+c=0$ to get $x^{2}-3x+1=0$, which gives $a=1$, $b=-3$, and $c=1$.
3. Calculate $a-b+c = 1 - (-3) + 1 = 1 + 3 + 1 = 5$.

**Answer:** (a) 5

> Common mistake: Forgetting to distribute the negative sign for $b$ when evaluating $a-b+c$.

### Question 32 (a)

*5 marks · Long answer*

The difference of the squares of two positive numbers is 180. The square of the smaller number is 8 times the greater number. Find the two numbers.

**Solution**

1. Let the greater number be $x$ and the smaller number be $y$.
2. According to the question, the square of the smaller number is $8$ times the greater number, so $y^2 = 8x$.
3. The difference of the squares of the two numbers is $180$, so $x^2 - y^2 = 180$.
4. Substitute $y^2 = 8x$ into the second equation to get $x^2 - 8x - 180 = 0$.
5. Factorise the quadratic equation: $x^2 - 18x + 10x - 180 = 0$.
6. Solve for $x$: $(x - 18)(x + 10) = 0$, which gives $x = 18$ or $x = -10$.
7. Since the numbers are positive, the greater number cannot be negative, so $x = 18$.
8. Find $y$: $y^2 = 8 \times 18 = 144$, which gives $y = \pm 12$. Since the numbers are positive, $y = 12$.
9. Thus, the two numbers are $18$ and $12$.

**Answer:** The two numbers are 18 and 12.

> Common mistake: Taking both positive and negative values for $y$ without checking the condition that the numbers are positive.

### Question 32 (OR)

*5 marks · Long answer*

Find the value(s) of k for which the equation $2x^{2}+kx+3=0$ has real and equal roots. Hence, find the roots of the equations so obtained.

**Solution**

1. Compare the given quadratic equation $2x^2 + kx + 3 = 0$ with the standard form $ax^2 + bx + c = 0$ to get $a = 2$, $b = k$, and $c = 3$.
2. The condition for a quadratic equation to have real and equal roots is that its discriminant $D$ must be equal to zero.
3. Substitute the values of $a$, $b$, and $c$ into the discriminant formula: $D = b^2 - 4ac = k^2 - 4(2)(3) = k^2 - 24$.
4. Set the discriminant to zero: $k^2 - 24 = 0$.
5. Solve for $k$ to get $k^2 = 24$, which gives $k = \pm\sqrt{24} = \pm 2\sqrt{6}$.
6. Substitute $k = 2\sqrt{6}$ into the equation to get $2x^2 + 2\sqrt{6}x + 3 = 0$.
7. Rewrite the equation as $(\sqrt{2}x + \sqrt{3})^2 = 0$, which gives the equal roots $x = -\frac{\sqrt{3}}{\sqrt{2}} = -\sqrt{\frac{3}{2}}$ and $x = -\sqrt{\frac{3}{2}}$.
8. Substitute $k = -2\sqrt{6}$ into the equation to get $2x^2 - 2\sqrt{6}x + 3 = 0$.
9. Rewrite the equation as $(\sqrt{2}x - \sqrt{3})^2 = 0$, which gives the equal roots $x = \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{3}{2}}$ and $x = \sqrt{\frac{3}{2}}$.

**Answer:** $k = \pm 2\sqrt{6}$; for $k = 2\sqrt{6}$, roots are $-\sqrt{\frac{3}{2}}, -\sqrt{\frac{3}{2}}$; for $k = -2\sqrt{6}$, roots are $\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}}$

> Common mistake: Forget to write both positive and negative values of $k$ when taking the square root.

## CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions

### Question 34

*5 marks · Long answer*

The perimeter of a right triangle is 60 cm and its hypotenuse is 25 cm. Find the lengths of other two sides of the triangle.

**Solution**

1. Let the lengths of the other two sides of the right triangle be $x$ cm and $y$ cm.
2. The perimeter is given as 60 cm, so $x + y + \text{hypotenuse} = 60$, which means $x + y + 25 = 60$, or $x + y = 35$, so $y = 35 - x$.
3. By Pythagoras theorem, $x^2 + y^2 = 25^2$, so $x^2 + (35 - x)^2 = 625$.
4. Expanding the equation, $x^2 + 1225 - 70x + x^2 = 625$, which simplifies to $2x^2 - 70x + 600 = 0$.
5. Dividing by 2, we get the quadratic equation $x^2 - 35x + 300 = 0$.
6. Factorising the quadratic equation, $(x - 20)(x - 15) = 0$, which gives $x = 20$ or $x = 15$.
7. If $x = 20$, then $y = 15$; if $x = 15$, then $y = 20$.
8. The lengths of the other two sides of the triangle are 15 cm and 20 cm.

**Answer:** 15 cm and 20 cm

> Common mistake: Errors in expanding $(35 - x)^2$ or solving the quadratic equation incorrectly.

### Question 34 (OR)

*5 marks · Long answer*

A train travels 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 speed of the train.

**Solution**

1. Let the uniform speed of the train be $x$ km/h.
2. Time taken to travel 480 km at speed $x$ is $\frac{480}{x}$ hours.
3. If the speed is $(x - 8)$ km/h, the time taken is $\frac{480}{x - 8}$ hours.
4. According to the question, $\frac{480}{x - 8} - \frac{480}{x} = 3$.
5. Taking LCM, $\frac{480x - 480(x - 8)}{x(x - 8)} = 3$, which simplifies to $\frac{3840}{x^2 - 8x} = 3$.
6. Cross-multiplying and dividing by 3, $x^2 - 8x = 1280$, so $x^2 - 8x - 1280 = 0$.
7. Factorising the quadratic equation, $x^2 - 40x + 32x - 1280 = 0$, giving $x(x - 40) + 32(x - 40) = 0$, or $(x - 40)(x + 32) = 0$.
8. Since speed cannot be negative, $x = 40$.
9. The speed of the train is 40 km/h.

**Answer:** 40 km/h

> Common mistake: Writing the time difference as $\frac{480}{x} - \frac{480}{x-8} = 3$ instead of the correct larger time minus smaller time.

## CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions

### Question 6

*1 mark · MCQ*

If the roots of quadratic equation $4x^2 - 5x + k = 0$ are real and equal, then value of $k$ is :

- $\frac{5}{4}$
- $\frac{25}{16}$
- $-\frac{5}{4}$
- $-\frac{25}{16}$

**Solution**

1. For real and equal roots, the discriminant must be zero: $b^2 - 4ac = 0$.
2. Substitute $a = 4$, $b = -5$, and $c = k$ into the equation: $(-5)^2 - 4(4)(k) = 0$.
3. Solving for $k$, we get $25 - 16k = 0$, which gives $k = \frac{25}{16}$.

**Answer:** (b) $\frac{25}{16}$

> Common mistake: Forgetting the negative sign while squaring $-5$ or missing the $4ac$ term.

## CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions

### Question 3

*1 mark · MCQ*

If the roots of equation $ax^2 + bx + c = 0, a \neq 0$ are real and equal, then which of the following relation is true ?

- $a = \frac{b^2}{c}$
- $b^2 = ac$
- $ac = \frac{b^2}{4}$
- $c = \frac{b^2}{a}$

**Solution**

1. For a quadratic equation $ax^2 + bx + c = 0$ with real and equal roots, the discriminant must be zero, so $b^2 - 4ac = 0$.
2. Rearranging this gives $b^2 = 4ac$, which can be written as $ac = \frac{b^2}{4}$.

**Answer:** (c) $ac = \frac{b^2}{4}$

> Common mistake: Students often forget the factor of 4 and mistakenly write $b^2 = ac$ or $ac = b^2$.

### Question 36

*4 marks · Case-based*

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.

**Part (i)**

1. Let the original side length of each square tile be $x$ units.
2. The area of one tile is $x^2$ square units, so the total floor area is $200x^2$.
3. When the side length is increased by $1$ unit, the new side length is $(x + 1)$ units and the area of one tile is $(x + 1)^2$ square units.
4. Since $128$ such tiles cover the same floor, the total floor area is also $128(x + 1)^2$.
5. Equating the two expressions for the floor area gives $200x^2 = 128(x + 1)^2$.

Answer (i): $200x^2 = 128(x + 1)^2$

**Part (ii)**

1. Start with the equation $200x^2 = 128(x + 1)^2$.
2. Divide both sides by $8$ to simplify: $25x^2 = 16(x^2 + 2x + 1)$.
3. Expand the right side: $25x^2 = 16x^2 + 32x + 16$.
4. Transpose all terms to the left-hand side: $25x^2 - 16x^2 - 32x - 16 = 0$.
5. Combine like terms to get the standard form: $9x^2 - 32x - 16 = 0$.

Answer (ii): $9x^2 - 32x - 16 = 0$

**Part (iii)(a)**

1. Consider the quadratic equation $9x^2 - 32x - 16 = 0$.
2. Split the middle term: $9x^2 - 36x + 4x - 16 = 0$.
3. Factor by grouping: $9x(x - 4) + 4(x - 4) = 0$.
4. $(9x + 4)(x - 4) = 0$.
5. Since $x$ is a side length, $x = 4$ or $x = -\frac{4}{9}$ (rejected). Thus $x = 4$.

Answer (iii)(a): $x = 4$

**Answer:** The side length of each tile is $4$ units.

> Common mistake: Taking negative value of side length.

## CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions

### Question 4

*1 mark · MCQ*

The discriminant of the quadratic equation $2x^2 - 5x - 3 = 0$ is

- $1$
- $49$
- $7$
- $19$

**Solution**

1. For the quadratic equation $2x^2 - 5x - 3 = 0$, $a = 2$, $b = -5$, and $c = -3$.
2. The discriminant $D = b^2 - 4ac = (-5)^2 - 4(2)(-3) = 25 + 24 = 49$.

**Answer:** (b) $49$

> Common mistake: Sign errors while substituting negative values of $b$ or $c$ into $b^2 - 4ac$.

### Question 32

*5 marks · Long answer*

(a) The diagonal of a rectangular field is $60\text{ m}$ more than the shorter side. If the longer side is $80\text{ m}$ more than the shorter side, find the length of the sides of the field.

**Solution**

1. Let the shorter side of the rectangular field be $x \text{ m}$.
2. Then the length of the diagonal is $(x + 60) \text{ m}$ and the longer side is $(x + 80) \text{ m}$.
3. In a rectangle, the sides and the diagonal form a right-angled triangle, so by Pythagoras theorem, $\text{shorter side}^2 + \text{longer side}^2 = \text{diagonal}^2$.
4. Substitute the expressions: $x^2 + (x + 80)^2 = (x + 60)^2$.
5. Expand the brackets: $x^2 + x^2 + 160x + 6400 = x^2 + 120x + 3600$.
6. Simplify the equation to standard quadratic form: $x^2 + 40x + 2800 = 0$.
7. Factorise the quadratic equation: $x^2 + 70x - 30x - 2800 = 0$ or $(x + 70)(x - 30) = 0$.
8. This gives $x = 30$ or $x = -70$. Since side length cannot be negative, $x = 30$.
9. Shorter side = $30 \text{ m}$ and longer side = $30 + 80 = 110 \text{ m}$.

**Answer:** Shorter side = 30 m, Longer side = 110 m

> Common mistake: Taking the diagonal and longer side expressions incorrectly or forgetting to reject the negative value of $x$.

### Question 32 (OR)

*5 marks · Long answer*

(b) The sum of the ages of a father and his son is $45$ years. Five years ago, the product of their ages (in years) was $124$. Determine their present age.

**Solution**

1. Let the present age of the son be $x$ years.
2. Since the sum of the ages of the father and his son is $45$ years, the present age of the father is $(45 - x)$ years.
3. Five years ago, the son's age was $(x - 5)$ years and the father's age was $(45 - x - 5) = (40 - x)$ years.
4. According to the question, the product of their ages $5$ years ago was $124$, so $(x - 5)(40 - x) = 124$.
5. Expand the equation: $40x - x^2 - 200 + 5x = 124$.
6. Simplify to quadratic form: $-x^2 + 45x - 324 = 0$, which gives $x^2 - 45x + 324 = 0$.
7. Factorise the quadratic equation: $x^2 - 36x - 9x + 324 = 0$ or $(x - 36)(x - 9) = 0$.
8. This gives $x = 36$ or $x = 9$. Since the son cannot be older than the father, $x = 9$.
9. Son's present age = $9$ years and father's present age = $45 - 9 = 36$ years.

**Answer:** Son's age = 9 years, Father's age = 36 years

> Common mistake: Wrong formulation of ages 5 years ago or incorrect factorisation of the quadratic equation.

## CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions

### Question 8

*1 mark · MCQ*

The least positive value of $k$, for which the quadratic equation $2x^2 + kx - 4 = 0$ has rational roots, is

- $\pm 2\sqrt{2}$
- 2
- $\pm 2$
- $\sqrt{2}$

**Solution**

1. For the quadratic equation $2x^2 + kx - 4 = 0$, the discriminant is $D = b^2 - 4ac = k^2 - 4(2)(-4) = k^2 + 32$.
2. Roots are rational when the discriminant is a non-zero positive square or zero, but for the least positive value of $k$ from the options, check $k = 2$: $D = 2^2 + 32 = 36$, which is a perfect square. Thus $k = 2$ gives rational roots.

**Answer:** (b) 2

> Common mistake: Students confuse rational roots with real roots and set discriminant strictly greater than zero instead of checking square values from options.

### Question 36 (I)

*1 mark · Case-based*

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 $18\text{ cm}$ long and $12\text{ cm}$ wide. Write an algebraic equation depicting the above information.

**Part (i)**

1. The original length of the photo is $18\text{ cm}$ and the original width is $12\text{ cm}$.
2. The original area of the photo is $18 \times 12 = 216\text{ cm}^2$.
3. When the length and width are increased by $x\text{ units}$, the new length is $18 + x$ and the new width is $12 + x$.
4. The area of the new photo is doubled, so it is $2 \times 216 = 432\text{ cm}^2$.
5. The algebraic equation depicting the information is $(18 + x)(12 + x) = 432$ or $(18 + x)(12 + x) = 2 \times 216$.

Answer (i): $(18 + x)(12 + x) = 432$

**Answer:** $(18 + x)(12 + x) = 2 \times (18 \times 12)$

> Common mistake: Students often forget to double the original area on the right-hand side of the equation.

### Question 36 (II)

*1 mark · Case-based*

Write the corresponding quadratic equation in standard form.

**Part (ii)**

1. Expand the equation from the previous part: $216 + 18x + 12x + x^2 = 432$.
2. Simplify and collect all terms on one side: $x^2 + 30x + 216 - 432 = 0$.
3. Write the quadratic equation in standard form: $x^2 + 30x - 216 = 0$.

Answer (ii): $x^2 + 30x - 216 = 0$

**Answer:** x^2 + 30x - 216 = 0

> Common mistake: Sign errors while transposing terms to form the standard quadratic equation.

### Question 36 (III)

*2 marks · Case-based*

What should be the new dimensions of the enlarged photo?

**Part (iii)**

1. Solve the quadratic equation $x^2 + 30x - 216 = 0$ by splitting the middle term: $x^2 + 36x - 6x - 216 = 0$.
2. Factor by grouping: $x(x + 36) - 6(x + 36) = 0$, which gives $(x - 6)(x + 36) = 0$.
3. Reject $x = -36$ as dimensions cannot be negative, so $x = 6$.
4. Calculate the new dimensions: Length = $18 + 6 = 30\text{ cm}$ and Width = $12 + 6 = 24\text{ cm}$.

Answer (iii): New length is $30\text{ cm}$ and new width is $24\text{ cm}$.

**Answer:** 30 cm by 24 cm

> Common mistake: Accepting the negative value of $x$ or failing to add $x$ back to the original dimensions.

### Question 36 (III) (OR)

*2 marks · Case-based*

Can any rational value of $x$ make the new area equal to $220\text{ cm}^2$?

**Part (iii) (OR)**

1. Set the new area equal to $220\text{ cm}^2$: $(18 + x)(12 + x) = 220$.
2. Simplify the equation: $216 + 30x + x^2 = 220$, which gives $x^2 + 30x - 4 = 0$.
3. Calculate the discriminant $D = b^2 - 4ac = (30)^2 - 4(1)(-4) = 900 + 16 = 916$.
4. Since $D > 0$, real roots exist, but solve for $x$: $x = \frac{-30 \pm \sqrt{916}}{2} = -15 \pm \sqrt{229}$. Since $229$ is not a perfect square, $x$ is irrational, hence no rational value of $x$ is possible.

Answer (iii) (OR): No rational value of $x$ exists.

**Answer:** No, because the discriminant is negative.

> Common mistake: Confusing real values with rational values.

## CBSE Class 10 Maths Basic Question Paper 2022 (Set 430/1/1) with Solutions

### Question 3 (a)

*2 marks · Short answer*

Find the nature of the roots of the quadratic equation $x^2 - 5x + 9 = 0$.

**Solution**

1. Compare the given quadratic equation $x^2 - 5x + 9 = 0$ with the standard form $ax^2 + bx + c = 0$ to get $a = 1$, $b = -5$, and $c = 9$.
2. Find the discriminant $D = b^2 - 4ac$.
3. Substitute the values: $D = (-5)^2 - 4(1)(9) = 25 - 36 = -11$.
4. Since $D < 0$, the given quadratic equation has no real roots.

**Answer:** No real roots

> Common mistake: Making sign errors while calculating the square of negative numbers in the discriminant.

### Question 3 (OR)

*2 marks · Short answer*

Write a quadratic equation with roots $-3$ and $5$.

**Solution**

1. Let the roots be $\alpha = -3$ and $\beta = 5$.
2. Find the sum of the roots: $\alpha + \beta = -3 + 5 = 2$.
3. Find the product of the roots: $\alpha \beta = (-3)(5) = -15$.
4. Use the standard quadratic equation form $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$ to get $x^2 - 2x - 15 = 0$.

**Answer:** $x^2 - 2x - 15 = 0$

> Common mistake: Reversing the sign of the sum of roots coefficient in the quadratic equation.

### Question 5

*2 marks · Short answer*

Solve the quadratic equation $2x^2 - 5x - 1 = 0$ for $x$.

**Solution**

1. Compare the given equation $2x^2 - 5x - 1 = 0$ with the standard form $ax^2 + bx + c = 0$ to get $a = 2$, $b = -5$, and $c = -1$.
2. Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the roots.
3. Substitute the values of $a$, $b$, and $c$ into the formula to get $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-1)}}{2(2)}$.
4. Simplify the expression to obtain $x = \frac{5 \pm \sqrt{25 + 8}}{4} = \frac{5 \pm \sqrt{33}}{4}$.

**Answer:** $x = \frac{5 \pm \sqrt{33}}{4}$

> Common mistake: Making sign errors while substituting negative values of $b$ and $c$ into the quadratic formula.

## CBSE Class 10 Maths Standard Question Paper 2022 (Set 30/1/1) with Solutions

### Question 3 (a)

*2 marks · Short answer*

Find the value of $m$ for which the quadratic equation $(m-1)x^2 + 2(m-1)x + 1 = 0$ has two real and equal roots.

**Solution**

1. For a quadratic equation $ax^2 + bx + c = 0$ to have real and equal roots, the discriminant must be zero: $D = b^2 - 4ac = 0$.
2. Here $a = m-1$, $b = 2(m-1)$, and $c = 1$.
3. Substitute into discriminant: $[2(m-1)]^2 - 4(m-1)(1) = 0$.
4. Simplify: $4(m-1)^2 - 4(m-1) = 0 \implies 4(m-1)(m-1-1) = 0 \implies 4(m-1)(m-2) = 0$.
5. Since $m-1 \neq 0$ for a quadratic equation (as coefficient of $x^2$ cannot be 0), we get $m = 2$.

**Answer:** $m = 2$

> Common mistake: Dividing out $(m-1)$ without checking if $m=1$ makes the equation linear.

### Question 3 (b) (OR)

*2 marks · Short answer*

Solve the following quadratic equation for $x$ : $\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0$.

**Solution**

1. Split the middle term for the equation $\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0$.
2. Find two numbers whose product is $\sqrt{3} \times 7\sqrt{3} = 21$ and sum is $10$. These are $7$ and $3$.
3. Rewrite the equation: $\sqrt{3}x^2 + 7x + 3x + 7\sqrt{3} = 0$.
4. Factor by grouping: $x(\sqrt{3}x + 7) + \sqrt{3}(\sqrt{3}x + 7) = 0 \implies (x + \sqrt{3})(\sqrt{3}x + 7) = 0$.
5. Solve for $x$: $x = -\sqrt{3}$ or $x = -\frac{7}{\sqrt{3}}$.

**Answer:** $x = -\sqrt{3}, -\frac{7}{\sqrt{3}}$

> Common mistake: Sign errors while grouping terms with radicals.

### Question 5

*2 marks · Short answer*

The product of Rehan's age (in years) $5$ years ago and his age $7$ years from now, is one more than twice his present age. Find his present age.

**Solution**

1. Let Rehan's present age be $x$ years.
2. His age 5 years ago was $(x - 5)$ years and his age 7 years from now is $(x + 7)$ years.
3. According to the given condition, $(x - 5)(x + 7) = 2x + 1$.
4. Expanding and simplifying, $x^2 + 2x - 35 = 2x + 1$, which gives $x^2 - 36 = 0$.
5. Solving for $x$, $x^2 = 36$, so $x = 6$ (since age cannot be negative).

**Answer:** 6 years

> Common mistake: Taking the product equal to $2x + 1$ incorrectly or forgetting to reject the negative value of age.

## Related pages

- [Quadratic Equations: NCERT solutions](https://www.swavid.com/maths/class/10/chapter/quadratic-equations/ncert-solutions)
- [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student