Updated: 09 Apr 2023

1810

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

1. The name Quadratic comes from
O Dratic
O Both a & b
O None of these
Explanation:

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 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 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 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

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 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 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

Explanation:
x^2-8 x+9=0
x^2-8 x=-9
x^2-8 x+(4)^2=-9+(4)^2
(x)^2-2(x)(4)+(4)^2=-9+16
(x-4)^2=7

25. By ________ we can solve all types of quadratic equations.
O Factorization method
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 None of these
Explanation:
Quadratic Formula is derived by Completing Square

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 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

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

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

1. Polynomial of degree four is called
O Both a & b
O None of these

Explanation:

2. The equation in the form of a x^4+b x^2+c=0 is called
O Both a & b
O None of these

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

O One
O Two
O Three
O 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

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 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 Both a & b
O None of these

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

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

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 Exponential
O All of these

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

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

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

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

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 Zero-zeroproperty
O None of these

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

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 kAll of these

Explanation:
These all are the steps to solve Exponential equation.

## Exercise # 1.3

1. An equation in which the variable appear in one or more radicands is called a equation.
O Linear
O All of these

Explanation:

2. In \sqrt{x+2}, \ x+2 is
O All of these

Explanation:

3. \sqrt{x+2}=3 is _______ equation.
O Linear
O All of these

Explanation:

4. Square root is finished by
O Formula
O Squaring
O None of these

Explanation:

5. The solution satisfies the original radical equation is called
O Solution set
O Extraneous
O Squaring

Explanation:

6. The solution that does not satisfy the original radical equation is called
O Solution set
O Extraneous
O Squaring

Explanation:

7. (\sqrt{x+2})^2=
O x^2+4
O x^2+4+2(x)(2)
O x^2
O x+2

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

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

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)}

O Identical (same)
O All
O Different
O Squaring

Explanation:

11. \sqrt{9}+\sqrt{16}=
O 3+4
O 7
O Both a & b
O None of these

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

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

Explanation:
\sqrt{9}+\sqrt{16}
=3+4
=7

14. \sqrt{9}+\sqrt{16} _______ \sqrt{9+16}
O Equal to
O Not equal to
O None of these

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

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

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
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}
\frac{1 \pm \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
cannot be Simplified
Explanation:

(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
a=2, b=-1, c=-3
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
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}
x= \frac{-2 \pm \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}
x =\pm \frac{1}{2}
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
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
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