Class 10 Maths Chapter 4 Quadratic Equations MCQs

Class 10 Maths MCQs for Chapter 4 (Quadratic Equations) are available online here with answers. All these objective questions are prepared as per the latest CBSE syllabus (2022 – 2023) and NCERT guidelines. MCQs for Class 10 Maths Chapter 4 are prepared according to the new exam pattern. Solving these multiple-choice questions will help students to score good marks in the board exams, which they can verify with the help of detailed explanations given here. To get chapter-wise MCQs, click here. Also, find the PDF of MCQs to download here for free.

Class 10 Maths MCQs for Quadratic Equations

CBSE board has released the datasheet for the Class 10 Maths exam. It is advised for students to start revising the chapters, for the exam. Here, we have given multiple-choice questions for Chapter 4 quadratic equations, to help students to solve different types of questions, which could appear in the board exam. They can build their problem-solving capacity and boost their confidence level by practising the questions here. Get important questions for class 10 Maths here at BYJU’S.

Students can also get access to Quadratic equations Class 10 Notes here.

Below are the MCQs for Quadratic Equations

1. Equation of (x+1)2-x2=0 has number of real roots equal to:

(a) 1

(b) 2

(c) 3

(d) 4

Explanation: (x+1)2-x2=0

X2+2x+1-x2 = 0

2x+1=0

x=-½

Hence, there is one real root.

2. The roots of 100x2 – 20x + 1 = 0 is:

(a) 1/20 and 1/20

(b) 1/10 and 1/20

(c) 1/10 and 1/10

(d) None of the above

Answer: (c) 1/10 and 1/10

Explanation: Given, 100x2 – 20x + 1=0

100x2 – 10x – 10x + 1 = 0

10x(10x – 1) -1(10x – 1) = 0

(10x – 1)2 = 0

∴ (10x – 1) = 0 or (10x – 1) = 0

⇒x = 1/10 or x = 1/10

3. The sum of two numbers is 27 and product is 182. The numbers are:

(a) 12 and 13

(b) 13 and 14

(c) 12 and 15

(d) 13 and 24

Answer: (b) 13 and 14

Explanation: Let x is one number

Another number = 27 – x

Product of two numbers = 182

x(27 – x) = 182

⇒ x2 – 27x – 182 = 0

⇒ x2 – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

⇒ x = 13 or x = 14

4. If ½ is a root of the quadratic equation x2-mx-5/4=0, then value of m is:

(a) 2

(b) -2

(c) -3

(d) 3

Explanation: Given x=½ as root of equation x2-mx-5/4=0.

(½)2 – m(½) – 5/4 = 0

¼-m/2-5/4=0

m=-2

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, the other two sides of the triangle are equal to:

(a) Base=10cm and Altitude=5cm

(b) Base=12cm and Altitude=5cm

(c) Base=14cm and Altitude=10cm

(d) Base=12cm and Altitude=10cm

Answer: (b) Base=12cm and Altitude=5cm

Explanation: Let the base be x cm.

Altitude = (x – 7) cm

In a right triangle,

Base2 + Altitude2 = Hypotenuse2 (From Pythagoras theorem)

∴ x2 + (x – 7)2 = 132

By solving the above equation, we get;

⇒ x = 12 or x = – 5

Since the side of the triangle cannot be negative.

Therefore, base = 12cm and altitude = 12-7 = 5cm

6. The roots of quadratic equation 2x2 + x + 4 = 0 are:

(a) Positive and negative

(b) Both Positive

(c) Both Negative

(d) No real roots

Answer: (d) No real roots

Explanation: 2x2 + x + 4 = 0

⇒ 2x2 + x = -4

Dividing the equation by 2, we get

⇒ x2 + 1/2x = -2

⇒ x2 + 2 × x × 1/4 = -2

By adding (1/4)2 to both sides of the equation, we get

⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = (1/4)2 – 2

⇒ (x + 1/4)2 = 1/16 – 2

⇒ (x + 1/4)2 = -31/16

The square root of negative number is imaginary, therefore, there is no real root for the given equation.

$$\begin{array}{l}\text{7. The value of }\sqrt{6+\sqrt{6+\sqrt{6} \ldots \ldots \ldots}} \text{ is: }\end{array}$$

(a) 4

(b) 3

(c) 3.5

(d) -3

Explanation:

$$\begin{array}{l}\text{Let, }\sqrt{6+\sqrt{6+\sqrt{6} \ldots \ldots \ldots}} = x\end{array}$$

Hence, we can write, √(6+x) = x

6+x=x2

x2-x-6=0

x2-3x+2x-6=0

x(x-3)+2(x-3)=0

(x+2) (x-3) = 0

x=-2,3

Since, x cannot be negative, therefore, x=3

8. The sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is 1/3. The present age of Rehman is:

(a) 7

(b) 10

(c) 5

(d) 6

Explanation: Let, x is the present age of Rehman

Three years ago his age = x – 3

Five years later his age = x + 5

Given, the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3.

∴ 1/x-3 + 1/x-5 = 1/3

(x+5+x-3)/(x-3)(x+5) = 1/3

(2x+2)/(x-3)(x+5) = 1/3

⇒ 3(2x + 2) = (x-3)(x+5)

⇒ 6x + 6 = x2 + 2x – 15

⇒ x2 – 4x – 21 = 0

⇒ x2 – 7x + 3x – 21 = 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0

⇒ x = 7, -3

We know age cannot be negative, hence the answer is 7.

9. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

(a) 30 km/hr

(b) 40 km/hr

(c) 50 km/hr

(d) 60 km/hr

Answer: (b) 40 km/hr

Explanation: Let x km/hr be the speed of train.

Time required to cover 360 km = 360/x hr.

As per the question given,

⇒ (x + 5)(360-1/x) = 360

⇒ 360 – x + 1800-5/x = 360

⇒ x2 + 5x + 10x – 1800 = 0

⇒ x(x + 45) -40(x + 45) = 0

⇒ (x + 45)(x – 40) = 0

⇒ x = 40, -45

Negative value is not considered for speed hence the answer is 40km/hr.

10. If one root of equation 4x2-2x+k-4=0 is reciprocal of the other. The value of k is:

(a) -8

(b) 8

(c) -4

(d) 4

Explanation: If one root is reciprocal of others, then the product of roots will be:

α x 1/α = (k-4)/4

k-4=4

k=8

11. Which one of the following is not a quadratic equation?

(a) (x + 2)2 = 2(x + 3)

(b) x2 + 3x = (–1) (1 – 3x)2

(c) (x + 2) (x – 1) = x2 – 2x – 3

(d) x3 – x2 + 2x + 1 = (x + 1)3

Answer: (c) (x + 2) (x – 1) = x2 – 2x – 3

Explanation:

We know that the degree of a quadratic equation is 2.

By verifying the options,

(a) (x + 2)2 = 2(x + 3)

x2 + 4x + 4 = 2x + 6

x2 + 2x – 2 = 0

This is a quadratic equation.

(b) x2 + 3x = (–1) (1 – 3x)2

x2 + 3x = -1(1 + 9x2 – 6x)

x2 + 3x + 1 + 9x2 – 6x = 0

10x2 – 3x + 1 = 0

This is a quadratic equation.

(c) (x + 2) (x – 1) = x2 – 2x – 3

x2 + x – 2 = x2 – 2x – 3

x2 + x – 2 – x2 + 2x + 3 = 0

3x + 1 = 0

This is not a quadratic equation.

12. Which of the following equations has 2 as a root?

(a) x2 – 4x + 5 = 0

(b) x2 + 3x – 12 = 0

(c) 2x2 – 7x + 6 = 0

(d) 3x2 – 6x – 2 = 0

Answer: (c) 2x2 – 7x + 6 = 0

Explanation:

If 2 is a root then substituting the value 2 in place of x should satisfy the equation.

Let us verify the given options.

(a) x2 – 4x + 5 = 0

(2)2 – 4(2) + 5 = 1 ≠ 0

So, x = 2 is not a root of x2 – 4x + 5 = 0

(b) x2 + 3x – 12 = 0

(2)2 + 3(2) – 12 = -2 ≠ 0

So, x = 2 is not a root of x2 + 3x – 12 = 0

(c) 2x2 – 7x + 6 = 0

2(2)2 – 7(2) + 6 = 0

Here, x = 2 is a root of 2x2 – 7x + 6 = 0

13. A quadratic equation ax2 + bx + c = 0 has no real roots, if

(a) b2 – 4ac > 0

(b) b2 – 4ac = 0

(c) b2 – 4ac < 0

(d) b2 – ac < 0

Answer: (c) b2 – 4ac < 0

A quadratic equation ax+ bx + c = 0 has no real roots, if b2 – 4ac < 0. That means, the quadratic equation contains imaginary roots.

14. The product of two consecutive positive integers is 360. To find the integers, this can be represented in the form of quadratic equation as

(a) x2 + x + 360 = 0

(b) x2 + x – 360 = 0

(c) 2x2 + x – 360

(d) x2 – 2x – 360 = 0

Answer: (b) x2 + x – 360 = 0

Explanation:

Let x and (x + 1) be the two consecutive integers.

According to the given,

x(x + 1) = 360

x2 + x = 360

x2 + x – 360

15. The equation which has the sum of its roots as 3 is

(a) 2x2 – 3x + 6 = 0

(b) –x2 + 3x – 3 = 0

(c) √2x2 – 3/√2x + 1 = 0

(d) 3x2 – 3x + 3 = 0

Answer: (b) –x2 + 3x – 3 = 0

Explanation:

The sum of the roots of a quadratic equation ax2 + bx + c = 0, a ≠ 0 is given by,

Coefficient of x / coefficient of x2 = –(b/a)

Let us verify the options.

(a) 2x2 – 3x + 6 = 0

Sum of the roots = – b/a = -(-3/2) = 3/2

(b) -x2 + 3x – 3 = 0

Sum of the roots = – b/a = -(3/-1) = 3

(c) √2x2 – 3/√2x + 1=0

2x2 – 3x + √2 = 0

Sum of the roots = – b/a = -(-3/2) = 3/2

(d) 3x2 – 3x + 3 = 0

Sum of the roots = – b/a = -(-3/3) = 1

16. The quadratic equation 2x2 – √5x + 1 = 0 has

(a) two distinct real roots

(b) two equal real roots

(c) no real roots

(d) more than 2 real roots

Answer: (c) no real roots

Explanation:

Given,

2x2 – √5x + 1 = 0

Comparing with the standard form of a quadratic equation,

a = 2, b = -√5, c = 1

Now,

b2 – 4ac = (-√5)2 – 4(2)(1)

= 5 – 8

= -3 < 0

Therefore, the given equation has no real roots.

17. The equation (x + 1)2 – 2(x + 1) = 0 has

(a) two real roots

(b) no real roots

(c) one real root

(d) two equal roots

Answer: (a) two real roots

Explanation:

(x + 1)2 – 2(x + 1) = 0

x2 + 1 + 2x – 2x – 2 = 0

x2 – 1 = 0

x2 = 1

x = ± 1

18. The quadratic formula to find the roots of a quadratic equation ax2 + bx + c = 0 is given by

(a) [-b ± √(b2-ac)]/2a

(b) [-b ± √(b2-2ac)]/a

(c) [-b ± √(b2-4ac)]/4a

(d) [-b ± √(b2-4ac)]/2a

Answer: (d) [-b ± √(b2-4ac)]/2a

The quadratic formula to find the roots of a quadratic equation ax2 + bx + c = 0 is given by [-b ± √(b2-4ac)]/2a.

19. The quadratic equation x2 + 7x – 60 has

(a) two equal roots

(b) two real and unequal roots

(b) no real roots

(c) two equal complex roots

Answer: (b) two real and unequal roots

Explanation:

Given,

x2 + 7x – 60 = 0

Comparing with the standard form,

a = 1, b = 7, c = -60

b2 – 4ac = (7)2 – 4(1)(-60) = 49 + 240 = 289 > 0

Therefore, the given quadratic equation has two real and unequal roots.

20. The maximum number of roots for a quadratic equation is equal to

(a) 1

(b) 2

(c) 3

(d) 4