Quadratic Equations — Class 10 Maths practice questions

Question 1

A farmer wants to fence a rectangular vegetable garden such that its length is 3 m more than twice its breadth. The area of the garden is 90 sq. m. He also notes that the perimeter of the garden will determine how much fencing wire he needs to purchase.

Accepted Answer

Let the breadth of the garden be x m.

∴ Length = (2x + 3) m.

(i)

According to the question, Area = length × breadth = 90

⟹ x(2x + 3) = 90

⟹ 2x<super>2</super> + 3x − 90 = 0

∴ The required quadratic equation is 2x<super>2</super> + 3x − 90 = 0.

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(ii)

For 2x<super>2</super> + 3x − 90 = 0, we have a = 2, b = 3, c = −90.

D = b<super>2</super> − 4ac

⟹ D = (3)<super>2</super> − 4(2)(−90)

⟹ D = 9 + 720

∴ D = 729

---

(iii)

Solving 2x<super>2</super> + 3x − 90 = 0 by splitting the middle term:

⟹ 2x<super>2</super> + 15x − 12x − 90 = 0

⟹ x(2x + 15) − 6(2x + 15) = 0

⟹ (x − 6)(2x + 15) = 0

⟹ x = 6 or x = −15/2

∵ breadth cannot be negative, x = −15/2 is rejected.

∴ Breadth = x = 6 m

∴ Length = 2(6) + 3 = 15 m

Perimeter (fencing wire required) = 2(l + b) = 2(15 + 6) = 2 × 21

∴ Fencing wire required = 42 m

Question 2

A school is organising a science fair and has booked a rectangular exhibition hall. The length of the hall is 3 m more than twice its breadth. A display partition divides the hall into two sections, and the total area of the hall is 90 sq. m. The organising committee needs to find the exact dimensions of the hall to plan the arrangement of stalls.

Accepted Answer

**(i)** Let the breadth of the hall be x metres. ⟹ Length = (2x + 3) metres. According to the question, x(2x + 3) = 90 ⟹ 2x2 + 3x − 90 = 0 **(ii)** For 2x2 + 3x − 90 = 0, a = 2, b = 3, c = −90. D = b2 − 4ac = (3)2 − 4(2)(−90) = 9 + 720 = 729 ∴ D = 729 > 0, so the equation has two distinct real roots. **(iii)** Using the quadratic formula, x = (−b ± √D) / 2a ⟹ x = (−3 ± √729) / (2 × 2) = (−3 ± 27) / 4 ⟹ x = (−3 + 27)/4 = 24/4 = 6 or x = (−3 − 27)/4 = −30/4 ∵ breadth cannot be negative, x = −30/4 is rejected. ∴ Breadth = **6 m** and Length = 2(6) + 3 = **15 m**.

Question 3

A group of friends planned a road trip from City A to City B, a total distance of 480 km. On the return journey, due to heavy traffic, their average speed decreased by 20 km/h compared to the onward journey, and as a result the return trip took 2 hours longer than the onward journey. Let the average speed during the onward journey be x km/h.

Accepted Answer

**(i)**

Time taken for onward journey = 480/x hours.

Time taken for return journey = 480/(x − 20) hours.

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**(ii)**

According to the question, the return journey took 2 hours longer than the onward journey:

480/(x − 20) − 480/x = 2

Taking LCM x(x − 20):

[480x − 480(x − 20)] / [x(x − 20)] = 2

⟹ [480x − 480x + 9600] / [x(x − 20)] = 2

⟹ 9600 / [x(x − 20)] = 2

Cross-multiplying:

9600 = 2x(x − 20)

⟹ 4800 = x² − 20x

⟹ **x² − 20x − 4800 = 0**

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**(iii)**

Solving x² − 20x − 4800 = 0 by factorisation.

Two numbers whose product = −4800 and sum = −20 are −80 and 60.

⟹ x² − 80x + 60x − 4800 = 0

⟹ x(x − 80) + 60(x − 80) = 0

⟹ (x − 80)(x + 60) = 0

⟹ x = 80 or x = −60

∵ speed cannot be negative, x = −60 is rejected.

∴ The average speed during the onward journey is **80 km/h**, and the average speed during the return journey is 80 − 20 = **60 km/h**.

Question 4

A rectangular swimming pool is being constructed in a community park. The length of the pool is 3 m more than twice its breadth. The area of the pool is 90 m². The construction team also plans to lay tiles along a uniform border of width 1 m around the pool, and they need to calculate various dimensions before ordering materials.

Accepted Answer

**(i)**

Let breadth = x metres.

⟹ Length = (2x + 3) metres.

According to the question, length × breadth = 90

⟹ x(2x + 3) = 90

⟹ 2x² + 3x − 90 = 0

∴ The quadratic equation formed is **2x² + 3x − 90 = 0**.

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**(ii)**

Solving 2x² + 3x − 90 = 0 by splitting the middle term:

⟹ 2x² + 15x − 12x − 90 = 0

⟹ x(2x + 15) − 6(2x + 15) = 0

⟹ (x − 6)(2x + 15) = 0

⟹ x = 6  or  x = −15/2

∵ breadth cannot be negative, x = −15/2 is rejected.

∴ Breadth of the pool = **6 m**.

---

**(iii)**

From part (ii), breadth = 6 m.

⟹ Length = 2(6) + 3 = **15 m**.

With a uniform border of width 1 m around the pool, the outer rectangle has:

⟹ Outer length = 15 + 2(1) = 17 m,  Outer breadth = 6 + 2(1) = 8 m.

Area of tiled border = Area of outer rectangle − Area of pool

⟹ = (17 × 8) − (15 × 6)

⟹ = 136 − 90

∴ Area of the tiled border = **46 m²**.

Question 5

A farmer wants to fence a rectangular vegetable plot in his field. The length of the plot is 3 metres more than twice its breadth. The area of the rectangular plot is 90 square metres. The farmer needs to find the dimensions of the plot to purchase the correct amount of fencing material.

Accepted Answer

**(i)** Let breadth = x metres. ∴ Length = (2x + 3) metres. According to the question, Area = Length × Breadth ⟹ x(2x + 3) = 90 ⟹ 2x2 + 3x − 90 = 0 ∴ The required quadratic equation is **2x2 + 3x − 90 = 0**. --- **(ii)** For the quadratic equation 2x2 + 3x − 90 = 0, we have a = 2, b = 3, c = −90. Discriminant D = b2 − 4ac ⟹ D = (3)2 − 4(2)(−90) ⟹ D = 9 + 720 ∴ **D = 729** --- **(iii)** Solving 2x2 + 3x − 90 = 0 by factorisation: ⟹ 2x2 + 15x − 12x − 90 = 0 ⟹ x(2x + 15) − 6(2x + 15) = 0 ⟹ (x − 6)(2x + 15) = 0 ⟹ x = 6 or x = −15/2 ∵ breadth cannot be negative, x = −15/2 is rejected. ∴ Breadth = 6 m and Length = 2(6) + 3 = **15 m**.

Question 6

A school garden club is designing a rectangular vegetable patch. The length of the patch is 3 metres more than twice its breadth. The area of the patch is 27 square metres.

(i) If the breadth of the patch is x metres, write the quadratic equation representing the above situation.
(ii) Find the discriminant of the equation obtained in part (i). What does its value tell you about the roots?
(iii) Find the dimensions (length and breadth) of the vegetable patch.
OR
(iii) A second patch is to be designed such that its area is also 27 m², but this time the length equals twice the breadth. Using the discriminant, determine whether such a rectangular patch with integer dimensions is possible.

Accepted Answer

(i) Let the breadth of the patch = x metres.
∴ Length = (2x + 3) metres.

According to the question,
x(2x + 3) = 27
⟹ 2x² + 3x = 27
⟹ 2x² + 3x − 27 = 0

∴ The required quadratic equation is 2x² + 3x − 27 = 0.                    [1 mark]

(ii) Comparing 2x² + 3x − 27 = 0 with ax² + bx + c = 0:
a = 2, b = 3, c = −27

Discriminant D = b² − 4ac
⟹ D = (3)² − 4(2)(−27)
⟹ D = 9 + 216
⟹ D = 225

∴ D = 225 > 0

Since D > 0, the equation has two distinct real roots, meaning the patch has a unique valid solution for its dimensions.                    [1 mark]

(iii) Solving 2x² + 3x − 27 = 0 by factorisation:

2x² + 3x − 27 = 0
⟹ 2x² + 9x − 6x − 27 = 0
⟹ x(2x + 9) − 3(2x + 9) = 0
⟹ (2x + 9)(x − 3) = 0

⟹ x = 3  or  x = −9/2

∵ Breadth cannot be negative, x = −9/2 is rejected.

∴ Breadth = x = 3 metres
∴ Length = 2(3) + 3 = 9 metres                    [2 marks]

OR

(iii) Let the breadth of the second patch = x metres.
∴ Length = 2x metres.

According to the question,
x(2x) = 27
⟹ 2x² = 27
⟹ 2x² − 27 = 0

Comparing with ax² + bx + c = 0:
a = 2, b = 0, c = −27

D = b² − 4ac = (0)² − 4(2)(−27) = 0 + 216 = 216

∴ D = 216 > 0, so two distinct real roots exist.

x = √(27/2) = 3√(3/2) = (3√6)/2, which is irrational.

∵ The dimensions are irrational (not integers), such a rectangular patch with integer dimensions is NOT possible.                    [2 marks]

Question 7

Arjun and his younger sister Priya are trying to figure out their ages using a fun number puzzle. Arjun's age is 5 years more than twice Priya's age. The product of their present ages is 75. Their mother, a mathematics teacher, turned this into a classroom activity to demonstrate quadratic equations.

Accepted Answer

**(i)** Let Priya's present age = x years. ⟹ Arjun's present age = (2x + 5) years. According to the question, product of their ages = 75: x(2x + 5) = 75 ⟹ 2x2 + 5x = 75 ⟹ 2x2 + 5x − 75 = 0 ∴ The required quadratic equation is **2x2 + 5x − 75 = 0**. --- **(ii)** For 2x2 + 5x − 75 = 0, a = 2, b = 5, c = −75. Discriminant D = b2 − 4ac ⟹ D = (5)2 − 4(2)(−75) ⟹ D = 25 + 600 ∴ **D = 625** --- **(iii)** Using the quadratic formula: x = (−b ± √D) / 2a ⟹ x = (−5 ± √625) / (2 × 2) ⟹ x = (−5 ± 25) / 4 The two roots are: x = (−5 + 25) / 4 = 20/4 = 5 **or** x = (−5 − 25) / 4 = −30/4 = −7.5 ∵ age cannot be negative, x = −7.5 is rejected. ⟹ Priya's age = 5 years. ⟹ Arjun's age = 2(5) + 5 = 15 years. ∴ **Priya's present age = 5 years and Arjun's present age = 15 years.**

Question 8

A train travels a distance of 360 km at a uniform speed. Had the speed been 15 km/h more, the journey would have taken 3 hours less. A railway engineer uses this information to determine the original speed of the train and plan better scheduling for future routes.

Accepted Answer

**(i)**

Let the original speed of the train be x km/h.

Time taken at original speed = 360/x hours.

Time taken at increased speed (x + 15) km/h = 360/(x + 15) hours.

According to the question,

360/x − 360/(x + 15) = 3

⟹ 360(x + 15) − 360x = 3x(x + 15)

⟹ 5400 = 3x² + 45x

⟹ x² + 15x − 1800 = 0

∴ The required quadratic equation is **x² + 15x − 1800 = 0**.

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**(ii)**

For x² + 15x − 1800 = 0, here a = 1, b = 15, c = −1800.

Discriminant D = b² − 4ac

⟹ D = (15)² − 4(1)(−1800)

⟹ D = 225 + 7200

∴ **D = 7425**

---

**(iii)**

Solving x² + 15x − 1800 = 0 by factorisation.

Two numbers whose product = −1800 and sum = 15 are 60 and −45.

⟹ x² + 60x − 45x − 1800 = 0

⟹ x(x + 60) − 45(x + 60) = 0

⟹ (x − 45)(x + 60) = 0

⟹ x = 45 or x = −60

∵ speed cannot be negative, x = −60 is rejected.

∴ The original speed of the train is **45 km/h**.

Question 9

For the quadratic equation 3x² + kx + 3 = 0 to have two real and equal roots, the value of k is:

Accepted Answer

Option (A) is correct.

**Explanation:** For a quadratic equation to have two real and equal roots, the discriminant D = b² − 4ac = 0. Here a = 3, b = k, c = 3, so k² − 4(3)(3) = 0 ⟹ k² = 36 ⟹ k = ±6.

Question 10

If the quadratic equation (k + 1)x² − 2(k − 1)x + 1 = 0 has equal roots, then the value of k is:

Accepted Answer

Option (A) is correct.

Explanation: For equal roots, D = 0, where D = b² − 4ac. Here a = (k + 1), b = −2(k − 1), c = 1. ⟹ [−2(k − 1)]² − 4(k + 1)(1) = 0 ⟹ 4(k − 1)² − 4(k + 1) = 0 ⟹ (k − 1)² = k + 1 ⟹ k² − 2k + 1 = k + 1 ⟹ k² − 3k = 0 ⟹ k(k − 3) = 0 ∴ k = 0 or k = 3.

Question 11

If x = −3 is a root of the quadratic equation 4x² + px + 3 = 0, then the value of p is:

Accepted Answer

Option (C) is correct.

**Explanation:** Since x = −3 is a root of 4x² + px + 3 = 0, it must satisfy the equation. Substituting x = −3: 4(−3)² + p(−3) + 3 = 0 ⟹ 36 − 3p + 3 = 0 ⟹ 39 = 3p ⟹ p = 13.

Question 12

A quadratic equation 2x² + kx + 3 = 0 has two equal roots. Find the value of k.

Accepted Answer

Option (C) is correct.

Explanation: For a quadratic equation to have two equal roots, discriminant D = b² − 4ac = 0. Here a = 2, b = k, c = 3 ⟹ k² − 4(2)(3) = 0 ⟹ k² = 24 ⟹ k = ±√24 = ±2√6.

Question 13

The least positive value of k for which the quadratic equation 5x² + kx − 20 = 0 has rational roots, is:

Accepted Answer

Option (D) is correct.

Explanation: For rational roots, discriminant D = b² − 4ac must be a perfect square. Here D = k² − 4(5)(−20) = k² + 400. Testing the options in increasing order: k = 5 ⟹ 25 + 400 = 425 (not a perfect square); k = 10 ⟹ 100 + 400 = 500 (not a perfect square); k = 15 ⟹ 225 + 400 = 625 = 25², which is a perfect square. ∴ the least positive value of k is **15**.

Question 14

If the discriminant of the quadratic equation 5x² − 6x + m = 0 is 16, then the value of m is:

Accepted Answer

Option (C) is correct.

**Explanation:** For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. Here a = 5, b = −6, c = m, so D = (−6)² − 4(5)(m) = 36 − 20m. Setting D = 16 ⟹ 36 − 20m = 16 ⟹ 20m = 20 ⟹ m = 1.

Question 15

Assertion (A): The quadratic equation x² + x + 1 = 0 has no real roots.
Reason (R): The discriminant of x² + x + 1 is -3, which is negative.

Accepted Answer

Option (a) is correct.

Explanation: For x² + x + 1 = 0, D = b² − 4ac = (1)² − 4(1)(1) = 1 − 4 = −3. Since D < 0, the equation has no real roots, so Assertion (A) is true. Reason (R) states D = −3, which is negative — this is mathematically correct, so R is also true. Since a negative discriminant is precisely the condition that guarantees no real roots, R is the correct explanation of A.

Question 16

Assertion (A): The quadratic equation x² + x + 1 = 0 has no real roots.
Reason (R): The discriminant of x² + x + 1 is -3, which is negative.

Accepted Answer

Option (a) is correct.

Explanation: For x² + x + 1 = 0, D = b² − 4ac = (1)² − 4(1)(1) = 1 − 4 = −3 < 0. Since D < 0, the equation has no real roots, so A is true. R states D = −3, which is negative — this is true and directly explains why A holds (negative discriminant ⟹ no real roots). ∴ Both A and R are true and R is the correct explanation of A.

Question 17

The roots of the quadratic equation 6x² − 7x + 3 = 0 are:

Accepted Answer

Option (C) is correct.

Explanation: D = b² − 4ac = (−7)² − 4(6)(3) = 49 − 72 = −23. Since D < 0, the roots are not real.

Question 18

Assertion (A): The quadratic equation x² + x + 1 = 0 has no real roots.
Reason (R): The discriminant of x² + x + 1 is -3, which is negative.

Accepted Answer

Option (a) is correct.

Explanation: D = b² − 4ac = (1)² − 4(1)(1) = 1 − 4 = −3. Since D < 0, the equation x² + x + 1 = 0 has no real roots, so Assertion (A) is true. Reason (R) correctly states D = −3 and identifies it as negative, which is precisely the criterion (D < 0 ⟹ no real roots), so R is the correct explanation of A.

Question 19

For the quadratic equation 2x² + kx + 8 = 0 to have real and equal roots, the value of k is:

Accepted Answer

Option (A) is correct.

**Explanation:** For a quadratic equation to have real and equal roots, the discriminant D = b² − 4ac = 0. Here a = 2, b = k, c = 8 ⟹ k² − 4(2)(8) = 0 ⟹ k² = 64 ∴ k = ±8.

Question 20

If the quadratic equation x² + kx + 4 = 0 has two real and distinct roots, then the value of k can be:

Accepted Answer

Option (C) is correct.

Explanation: For two real and distinct roots, discriminant D = b² − 4ac > 0. Here a = 1, b = k, c = 4, so D = k² − 16 > 0 ⟹ k² > 16 ⟹ |k| > 4. Checking: k = 4 gives D = 0 (equal roots); k = −4 gives D = 0 (equal roots); k = 2 gives D = 4 − 16 = −12 < 0 (no real roots); k = 6 gives D = 36 − 16 = 20 > 0 ∴ two real and distinct roots.

Quadratic Equations: Class 10 Maths Practice Questions

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