# Quadratic Equation MCQs

Updated: 09 Apr 2023

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Quadratic equation MCQs can be challenging for students, but they can become much easier with the help of multiple-choice questions and detailed explanations. Whether you’re a student looking to improve your understanding or an educator searching for resources to help your students, these MCQs are a great place to start.

## Exercise # 1.1

### Quadratic Equation MCQs

**1. The name Quadratic comes from**

O Quad

O Dratic

O Both a & b

O None of these

Explanation:

**2. The word “quad” means**

O Cube

O Cubed root

O Square

O Square root

Explanation:

**3. An equation of degree is called quadratic equation.**

O 1

O 2

O 3

O 4

Explanation:

The name Quadratic comes from “quad” means square because the highest power of the variable is 2

**4. In quadratic equation a x^2+bx+c=0 ,**

O a=0

O a \neq 0

O Both a & b

O None of these

Explanation:

a=0 then it becomes linear equation.

**5. An equation of degree 2 is called equation.**

O Linear

O Quadratic

O Cubic

O All of these

Explanation:

a x^2+bx+c=0 is called General or Standard form of Quadratic equation.

**6. An equation of degree 1 is called ______ equation.**

O Linear

O Quadratic

O Cubic

O All of these

Explanation:

**7. In quadratic equation a x^2+b c+c=0, \ when \ a=0 then it becomes**

O Linear

O Quadratic

O Cubic

O All of these

Explanation:

a x^2+b c+c=0

when a=0

0x^2+b c+c=0

b c+c=0

This is linear equation.

**8. All those values of the variable for which the given equation is true are called**

O Solutions

O Roots

O Both a & b

O None of these

Explanation:

**9. The maximum number of roots of quadratic equation are**

O One

O Two

O Three

O All of these

Explanation:

The name Quadratic comes from “quad” means square because the highest power of the variable is 2.

**10. Which of the following is not a quadratic equation?**

O x^2+3 x+9=0

O x^2-16=0

O 9+3 x+x^2=0

O x^2+3 x^3+9=0

Explanation:

For Quadratic equation, the highest power of the variable is 2.

**11. There are _______ basic methods to solve Quadratic equation.**

O 1

O 2

O 3

O 5

Explanation:

1. Factorization

2. Completing square

3. Quadratic Formula

**12. In factorization method, a quadratic equation can be solved by _________ it in factors.**

O Combine

O Separate

O Splitting

O All of these

Explanation:

In factorization method, the middle term of a quadratic equation can be Splitted.

**13. To solve quadratic equation, the equation must have in ________ form of quadratic equation.**

O Any

O Standard

O Linear form

O All of these

Explanation:

The Standard form of quadratic equation is:

a x^2+b x+c=0

### By Factorization Method MCQs

**14. In factorization method, the ________ term will be split of a x^2+b x+c=0**

O a

O b

O c

O All of these

Explanation:

In factorization, the middle term should be split.

**15. In factorization method of a x^2+b x+c=0**

O We find the product of a \ (coefficient \ of \ x^2 ) \ and \ c \ (constant \ term) \ i.e. \ a c

O Find two numbers b_1 \ and \ b_2 such that b_1 \pm b_2=b \ and \ also \ b_1 b_2=a \cdot c

O a x^2+b_1 x+b_2 x+c=0 can be factorized into two limear factors.

O All of these

Explanation:

These all are the steps to solve Quadratic equation by Factorization.

**16. In factorization method, put all the terms on one side and _____ on other side.**

O 0

O 1

O B

O c

Explanation:

because Equate each factor to zero by zero – product property.

**17. In factorization method, equate each factor to ______ by zero- product property.**

O 0

O 1

O Constant

O Variable

Explanation:

In factorization method, put all the terms on one side and 0 on other side.

**18. In factorization method, equate each factor to zero by ________ property.**

O Quadratic-product

O Zero-product

O Both a & b

O None of these

Explanation:

In factorization method, put all the terms on one side and 0 on other side.

**19. In zero-product, if a b=0 , then either a=0 \ or \ b ______ 0**

O =

O \neq

O Both a & b

O None of these

Explanation:

if a b=0 , it means at least one must be zero either a or b.

**20. The solution of p^2+p-6=0 is**

O p=2,-3

O p=-2,-3

O p=2,3

O p=-2,3

Explanation:

p^2+p-6=0

p^2-2p+3p-6=0

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

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

p-2=0 \ or \ p+3=0

p=2 \ or \ p=-3

### By Completing Square MCQs

**21. Quadratic equation which cannot be solved by factorization, then it will be solved by**

O Completing square

O Quadratic Formula

O Both a & b

O None of these

Explanation:

By Completing Square & Quadratic Formula, we can solve almost every Quadratic Equation.

**22. In completing square method, the co-efficient of x^2 should be**

O 0

O 1

O Other than 1

O All of these

Explanation:

It is the first rule to solve Qudratic Equation by Completing Square.

**23. In completing square method of a x^2+b x+c=0**

O Divide all terms by the co-efficient of x^2 if other than 1

O Shift the constant term to the right side of the equation.

O Multiply the co-efficient of x \ with \ \frac{1}{2} then take Square of it and Add to B.S

O All of these

Explanation:

These all are the steps to solve Quadratic equation by Completing Square.

**24. To solve x^2-8 x+9=0 by completing square, then it becomes**

O (x-4)^2=7

O \sqrt{x-4}=7

O Both a & b

O None of these

Answer: (x-4)^2=7

Explanation:

x^2-8 x+9=0

x^2-8 x=-9

Add (4)^2 on B.S

x^2-8 x+(4)^2=-9+(4)^2

(x)^2-2(x)(4)+(4)^2=-9+16

(x-4)^2=7

### By Quadratic Formula MCQs

**25. By ________ we can solve all types of quadratic equations.**

O Factorization method

O Quadratic formula

O Both a & b

O None of these

Explanation:

**26. To solve a x^2+b x+c=0 by completing square, we get**

O Factors

O Quadratic formula

O Bi-quadratic

O None of these

Explanation:

Quadratic Formula is derived by Completing Square

**27. The quadratic formula is**

O x=\frac{-b }{2 a} \pm\sqrt{b^2-4 a c}

O x=\frac{-b \pm \sqrt{b^2}}{2 a}-4 a c

O x=-b \pm \frac{\sqrt{b^2-4 a c}}{2 a}

O x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}

Answer: x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}

**28. To apply Quadratic formula to 3 x^2-6 x+2=0 then**

O a=3, b=6, c=-2

O a=3, b=6, c=2

O a=3, b=-6, c=-2

O a=3, b=-6, c=2

Explanation:

Compare 3 x^2-6 x+2=0 with

a x^2+b x+c=0

**29. The solution set of 4 x^2+12 x=0 is**

O Solution \ Set =\{3\}

O { S.S }=\{0,3\}

O { S.S }=\{0,-3\}

O { S.S }=\{-3\}

Explanation:

4 x^2+12 x=0

4x(x+3)=0

4x=0 \ or \ x+3=0

x=\frac{0}{4} \ or \ x=-3

x=0 \ or \ x=-3

S.S=\{0, -3\}

**30. By ________ we can solve all quadratic equations.**

O Factorization method

O Quadratic formula

O Both a & b

O None of these

Explanation:

**31. The solution set of x^2+5 x+4=0 is**

O Solution \ Set =\{-1\}

O { S.S }=\{1,4\}

O { S.S }=\{-1,-4\}

O { S.S }=\{-4\}

Explanation:

x^2+5 x+4=0

x^2+1x+4x+4=0

x(x+1)+4(x+1)=0

(x+1)(x+4)=0

x+1=0 \ or \ x+4=0

x=-1 \ or \ x=-4

S.S=\{-1, -4\}

**32. The solution set of (x-3)^2=4 is**

O Solution \ Set =\{1\}

O S.S=\{1,5\}

O S.S=\{-1,-4\}

O S.S =\{-4\}

Explanation:

(x-3)^2=4

Taking Square root on B.S

\sqrt{(x-3)^2}=\pm \sqrt{4}

x-3=\pm 2

x-3=2 \ or \ x-3=-2

x=2+3 \ or \ x=-2+3

x=5 \ or \ x=1

S.S=\{5, 1\}

**33. What must be added to x^2+5 x to obtain a perfect square?**

O \left(\frac{5}{2}\right)^2

O \frac{5}{2}

O 5

O 2

Answer: \left(\frac{5}{2}\right)^2

Explanation:

Multiply the co – efficient of x \ with \ \frac{1}{2} then take Square of it and Add to B.S.

34. What must be added to q^2-4 q to obtain a perfect square?

O (2)^2

O \frac{5}{2}

O 5

O 2

Answer: (2)^2

Explanation:

Multiply the co – efficient of x \ with \ \frac{1}{2} then take Square of it and Add to B.S.

4 \times \frac{1}{2}=2

(2)^2

## Exercise # 1.2

### Biquadratic

**1. Polynomial of degree four is called**

O Quadratic

O Biquadratic

O Both a & b

O None of these

Answer: Biquadratic

Explanation:

**2. The equation in the form of a x^4+b x^2+c=0 is called**

O Quadratic

O Biquadratic

O Both a & b

O None of these

Answer: Biquadratic

Explanation:

Here the highest power is 4, that is why it is called Biquadratic

**3. Biquadratic equation has solutions.**

O One

O Two

O Three

O Four

Answer: Four

Explanation:

Here the highest power is 4, so there will be 4 roots of Biquadratic equation.

**4. The equation a x^4+b x^2+c=0 has solutions.**

O One

O Two

O Three

O Four

Answer: Four

Explanation:

Here the highest power is 4, so there will be 4 roots of Biquadratic equation.

**5. To solve a x^4+b x^2+c=0**

O a\left(x^2\right)^2+b x^2+c=0

O Substitute y=x^2

O To make Quadratic Equation

O All of these

Answer: All of these

Explanation:

These all are the steps of to solve the above equation.

**6. The equation a x^4+b x^2+c=0 can be solved by reducing it into**

O Quadratic

O Biquadratic

O Both a & b

O None of these

Answer: Quadratic

Explanation:

It is to easy to solve Biquadratic equation by reducing it into Quadratic equation.

**7. In substitutional it must remember to go back and express the answers in terms of______ the variable.**

O New

O Original

O Both a & b

O None of these

Answer: Original

Explanation:

**8. To solve a\left(x^2+\frac{1}{x^2}\right)+b\left(x+\frac{1}{x}\right)+c=0, we substitute**

O x+\frac{1}{x}=y

O x^2+\frac{1}{x^2}=y^2-2

O Both a & b

O None of these

Answer: Both a & b

Explanation:

Let x+\frac{1}{x}=y

Taking square root on B.S

\left(x+\frac{1}{x} \right)^2=y^2

x^2+\frac{1}{x^2}+2=y^2

x^2+\frac{1}{x^2}=y^2-2

### Exponential Equations

9. Exponential involving the term a^x is called equations.O Radical

O Quadratic

O Exponential

O All of these

Answer: Exponential

Explanation:

**10. For exponential equation, a^x it must be noted that**

O a>0

O a \neq 1

O a=2

O Both a & b

Answer: Both a & b

Explanation:

rules to represent the exponential equation.

**11. In equation, 4.2^{2 x}-10.2^x+4=0 , substitute**

O 2^{2 x}=y

O 2^x=y

O x=y

O All of these

Answer: 2^x=y

Explanation:

It is the simplest way to convert exponential equation to Quadratic equation.

**12. If 2^x=2^3 , then**

O 2=2

O x \neq 3

O x=3

O All of these

Answer: x=3

Explanation:

Here the bases are same and it is called One-to-One Property of Exponential Functions.

**13. If b^n=b^m , then**

O b=m

O n \neq m

O n=m

O All of these

Answer: n=m

Explanation:

Here the bases are same and it is called One-to-One Property of Exponential Functions.

**14. If b^n=b^m , then n=m is called ________ of exponential functions.**

O One-to-one property

O Quadratic property

O Zero-zeroproperty

O None of these

Answer: One-to-one property

Explanation:

Here the bases are same and it is called One-to-One Property of Exponential Functions.

**15. To solve 4x^2-10 x+4=0 , first we**

O Solve by factorization

O Taking 2 common

O Both a & b

O None of these

Answer: Taking 2 common

Explanation:

To make the equation in the simplest way.

**16. To solve 4.2^{2 x}-10.2^x+4=0**

O 4 .\left(2^x\right)^2-10.2^x+4=0

O 2^x=y

O To make Quadratic Equation

O kAll of these

Answer: All of these

Explanation:

These all are the steps to solve Exponential equation.

## Exercise # 1.3

### Radical Equations

**1. An equation in which the variable appear in one or more radicands is called a equation.**

O Radical

O Quadratic

O Linear

O All of these

Answer: Radical

Explanation:

**2. In \sqrt{x+2}, \ x+2 is**

O Radical

O Quadratic

O Radicand

O All of these

Answer: Radicand

Explanation:

**3. \sqrt{x+2}=3 is _______ equation.**

O Radical

O Quadratic

O Linear

O All of these

Answer: Radical

Explanation:

**4. Square root is finished by**

O Quadratic equation

O Formula

O Squaring

O None of these

Answer: Squaring

Explanation:

**5. The solution satisfies the original radical equation is called**

O Solution set

O Extraneous

O Quadratic

O Squaring

Answer: Solution Set

Explanation:

**6. The solution that does not satisfy the original radical equation is called**

O Solution set

O Extraneous

O Quadratic

O Squaring

Answer: Extraneous

Explanation:

**7. (\sqrt{x+2})^2=**

O x^2+4

O x^2+4+2(x)(2)

O x^2

O x+2

Answer: x+2

Explanation:

(\sqrt{x+2})^2= x+2

**8 . \quad(x+2)^2=**

O x^2+4+2(x)(2)

O x^2+4+4 x

O x^2+4

O Both a & b

Answer: Both a & b

Explanation:

(x+2)^2

=x^2+4+2(x)(2)

=x^2+4+4 x

**9. (\sqrt{x+2}+\sqrt{x+7})^2=**

O x+2+x+7

O x+2+x+7+\sqrt{(x+2)(x+7)}

O 2 x+9+2 \sqrt{(x+2)(x+7)}

O (\sqrt{(x+2)(x+7)})^2

Answer:

2 x+9+2 \sqrt{(x+2)(x+7)}

Explanation:

(\sqrt{x+2}+\sqrt{x+7})^2

=(\sqrt{x+2}+\sqrt{x+7})^2

=(\sqrt{x+2})^2+(\sqrt{x+7})^2+2\sqrt{x+2}\sqrt{x+7}

=x+2+x+7+2\sqrt{(x+2)(x+7)}

=2x+9+2\sqrt{(x+2)(x+7)}

**10 Addition of radicals is possible only with radical forms.**

O Identical (same)

O All

O Different

O Squaring

Answer: Identical (same)

Explanation:

**11. \sqrt{9}+\sqrt{16}=**

O 3+4

O 7

O Both a & b

O None of these

Answer: Both a & b

Explanation:

\sqrt{9}+\sqrt{16}

=3+4

=7

**12. \sqrt{9+16}=**

O \sqrt{25}

O 5

O Both a & b

O None of these

Answer: Both a & b

Explanation:

\sqrt{9+16}

=\sqrt{25}

=5

**13. \sqrt{9}+\sqrt{16}=**

O \sqrt{9+16}

O 7

O Both a & b

O None of these

Answer: 7

Explanation:

\sqrt{9}+\sqrt{16}

=3+4

=7

**14. \sqrt{9}+\sqrt{16} _______ \sqrt{9+16}**

O Equal to

O Not equal to

O Can add

O None of these

Answer: Not equal to

Explanation:

\sqrt{9}+\sqrt{16}

=3+4

=7

And

\sqrt{9+16}

=\sqrt{25}

=5

**15. If x^2=9 \ then \ x=**

O 3

O -3

O Both a & b

O None of these

Answer: Both a & b

Explanation:

x^2=9

\sqrt{x^2}=\pm \sqrt{9}

x=\pm 3

x=3 \ or \ x=-3

**16. If x^2=11 \ then \ x=**

O \pm \sqrt{11}

O \sqrt{11}

O -\sqrt{11}

O None of these

Answer: \pm \sqrt{11}

Explanation:

x^2=11

\sqrt{x^2}=\pm \sqrt{11}

## Review Exercise # 1

**(i). If (x+1)(x-5)=0 then the solutions are**

O x=1, -5

O x=1, 5

O x=-1, -5

O x=-1, 5

Explanation:

(x+1)(x-5)=0

x+1=o or x-5=0

x=-1 or x=5

Thus x=-1, 5

**(ii). if x^2-x-1=0 , then x=**

O \frac{-1 \pm \sqrt{5}}{2}

O -1 \pm \frac{\sqrt{5}}{2}

O \frac{1 \pm \sqrt{5}}{2}

O 1 \pm \frac{\sqrt{5}}{2}

Explanation:

x^2-x-1=0

a=1, b=-1, c=-1

We have

x= \frac{-b \pm \sqrt{b^2-4ac}}{2}

x= \frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-1)}}{2}

x= \frac{1 \pm \sqrt{1+4}}{2}

x= \frac{1 \pm \sqrt{5}}{2}

**(iii). \frac{-1 \pm \sqrt{5}}{2} in simplified form is**

O 1 \pm \sqrt{24}

O 1 \pm \sqrt{6}

O 2 \pm \sqrt{6}

O cannot be simplified

Explanation:

Already in simplified form

**(iv). To apply the quadratic formula to 2x^2-x=3**

a=2, b=-1, c=3

a=2, b=1, c=3

a=2, b=-1, c=-3

a=2, b=-1, c=0

Explanation:

2x^2-x=3

2x^2-x-3=0

Compare the equation with

ax^2+bx+c=0

a=2, b=-1, c=-3

**(v). If x^2-3x-4=0 , then the solutions are**

O x=4, -1

O x=-4, 1

O x=4, 1

O x=-4, -1

Explanation:

x^2-3x-4=0

x^2-4x+1x-4=0

x(x-4)+1(x-4)=0

(x-4)(x+1)=0

x-4=0 or x+1=0

x=4 or x=-1

**(vi). If 2x^2+4x-9=0 , then solutions are**

O x=\frac{2 \pm \sqrt{22}}{2}

O x=\frac{-2 \pm \sqrt{22}}{2}

O x=2 \pm \frac{\sqrt{22}}{2}

O x=-2 \pm \frac{\sqrt{22}}{2}

Explanation:

2x^2+4x-9=0

a=2, b=4, c=-9

We have

x= \frac{-b \pm \sqrt{b^2-4ac}}{2}

x= \frac{-4 \pm \sqrt{(4)^2-4(2)(-9)}}{(2)(2)}

x= \frac{-4 \pm \sqrt{16+72}}{4}

x= \frac{-4 \pm \sqrt{88}}{4}

x= \frac{-4 \pm \sqrt{4 \times 22}}{4}

x= \frac{-4 \pm 2\sqrt{22}}{4}

x= \frac{2(-2 \pm \sqrt{22})}{4}

x= \frac{-2 \pm \sqrt{22}}{2}

**(vii). x^2 - \frac{1}{4}=0 , then solution are:**

O x= \pm \frac{1}{2}

O x= \pm \frac{1}{4}

O x= \pm \frac{1}{8}

O x= \pm \frac{1}{16}

Explanation:

x^2 - \frac{1}{4}=0

x^2 = \frac{1}{4}

\sqrt {x^2} =\pm \sqrt{ \frac{1}{4}}

x =\pm \frac{1}{2}

**(viii). What are the solutions of the equation x^2+7x-18=0 ?**

O 2 or -9

O -2 or 9

O -2 or -9

O 2 or 9

Explanation:

x^2+7x-18=0

x^2-2x+9x-18=0

x(x-2)+9(x-2)=0

(x-2)(x+9)=0

x-2=0 or x+9=0

x=2 or x=-9

**(ix). Which of the following values of x are the roots of the equation x^2-8x+15=0 ?**

O x=1 or x=-7

O x=2 or x=4

O x=-2 or x=4

O x=3 or x=5

Explanation:

x^2-8x+15=0

x^2-3x-5x+15=0

x(x-3)-5(x-3)=0

(x-3)(x-5)=0

x-3=0 or x-5=0

x=3 or x=5

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