 Hello Aspirants, As we all know that Quadratic Equation Questions is a important part of quantitative aptitude section for every competitive exams.Different kind of Quadratic Equation Questions asked in every competitative exam.So here, In this article we will provide Quadratic Equation Quiz.These Quadratic Equation Questions are important for Bank, SSC, SEBI, NABARD, RBI, LIC, and Other state exams. You can attempt these questions & boost your preparation for your examination.

In the Banking exams Quadratic Equation Questions asked in the Prelims as well as Mains exam.There are 5 Quadratic Equation Questions asked in the Prelims exam (Bank).You want to score more in the Quadratic Equation section then you should practice more and more Quadratic Equation.

## Quadratic Equation Questions Quiz -3

Directions:(1-5) In the following questions, two equations are given.You have to solve both the equations and choose the correct option-

1. I. 9x2+17x+8=0
II. y2+8y+7=0
A. x ≤ y
B. x ≥ y
C. x < y
D. x > y
E. x = y or no specific relation can be established

Correct Answer – B. x ≥ y

First equation –
9x²+17x+8=0
=>9x²+9x+8x+8=0
=>9x(x+1)+8(x+1)=0
=>(9x+8)(x+1)=0
Thus we get, x = -8/9 or -1
Second equation –
y²+8y+7=0
=>y²+y+7y+7=0
=>y(y+1)+7(y+1)=0
=>(y+7)(y+1)=0
Thus we get, y = -7 or -1
Hence x ≥ y

2. I. x2-10x+24=0
II. 3y2+4y-15=0
A. x ≤ y
B. x ≥ y
C. x < y
D. x > y
E. x = y or no specific relation can be established

Correct Answer – D. x > y

First equation –
x²-10x+24=0
=>x²-4x-6x+24=0
=>x(x-4)-6(x-4)=0
=>(x-6)(x-4)=0
Thus we get, x = 6 or 4
Second equation –
3y²+4y-15=0
=>3y²+9y-5y-15=0
=>3y(y+3)-5(y+3)=0
=>(3y-5)(y+3)=0
Thus we get, y = 5/3 or -3
Hence x > y

3. I. x2+x-20=0
II. 9y2-55y+6=0
A. x ≤ y
B. x ≥ y
C. x < y
D. x > y
E. x = y or no specific relation can be established

Correct Answer – E. x = y or no specific relation can be established

First equation –
x²+x-20=0
=>x²+5x-4x-20=0
=>x(x+5)-4(x+5)=0
=>(x-4)(x+5)=0
Thus we get, x = 4 or -5
Second equation –
9y²-55y+6=0
=>9y²-54y-y+6=0
=>9y(y-6)-1(y-6)=0
=>(9y-1)(y-6)=0
Thus we get, y = 1/9 or 6
One value of x lies between the two values of y and vice versa, hence the relation cannot be determined.

4. I. x2-2x-24=0
II. y2+4y-45=0
A. x ≤ y
B. x ≥ y
C. x < y
D. x > y
E. x = y or no specific relation can be established

Correct Answer – E. x = y or no specific relation can be established

First equation –
x²-2x-24=0
=>x²-6x+4x-24=0
=>x(x-6)+4(x-6)=0
=>(x+4)(x-6)=0
Thus we get, x = -4 or 6
Second equation –
y²+4y-45=0
=>y²-5y+9y-45=0
=>y(y-5)+9(y-5)=0
=>(y+9)(y-5)=0
Thus we get, y = -9 or 5
One value of x lies between the two values of y and vice versa, hence the relation cannot be determined

5. I. x2+9x+8=0
II. 3y2+19y-14=0
A. x ≤ y
B. x ≥ y
C. x < y
D. x > y
E. x = y or no specific relation can be established

Correct Answer – E. x = y or no specific relation can be established

First equation –
x²+9x+8=0
=>x²+8x+x+8=0
=>x(x+8)+1(x+8)=0
=>(x+1)(x+8)=0
Thus we get, x = -1 or -8
Second equation –
3y²+19y-14=0
=>3y²+21y-2y-14=0
=>3y(y+7)-2(y+7)=0
=>(3y-2)(y+7)=0
Thus we get, y = 2/3 or -7
One value of x lies between the two values of y and vice versa, hence the relation cannot be determined

6.Based on the equations given choose the option which is true –

I. 10x2-31x-14=0
II. 5y2-8y-4=0
A. x ≤ y
B. x ≥ y
C. x < y
D. x > y
E. x = y or no specific relation can be established

Correct Answer – E. If x = y or no specific relation can be established

First equation –
10x2– 31x – 14 =0
or, 10x2– 35x + 4x – 14 = 0
or, 5x (2x – 7) + 2 (2x – 7) = 0
or, (5x + 2) (2x – 7) = 0
or, x = – 2/5 or 7/2
Second equation –
5y2– 8y – 4 = 0
or, 5y2– 10y + 2y – 4 = 0
or, 5y (y – 2) + 2 (y – 2) = 0
or, (5y + 2) (y – 2) = 0
or, y = – 2/5 or 2
Now here x can be more than y, it can be equal to y or it can be less than y. So the answer is 5

7.Based on the equations given choose the option which is true –

I. 4x2-1=0
II. 2y2+5y-3=0
A. x ≤ y
B. x ≥ y
C. x < y
D. x > y
E. x = y or no specific relation can be established

Correct Answer – E. If x = y or no specific relation can be established

First equation –
4x2– 1 = 0
or, x2 = ¼
so, x = ½ or – ½
Second equation –
2y2 + 5y – 3 = 0
or, 2y2 + 6y – y – 3 = 0
or, 2y (y + 3) – 1 (y + 3) = 0
or, (2y – 1) (y + 3) = 0
or, y = ½ or – 3
Here x can be more, less or equal to y so the answer is 5.

8.Based on the equations given choose the option which is true –

I. 3x2-11x+6=0
II. 2y2-3y-9=0
A. x ≤ y
B. x ≥ y
C. x < y
D. x > y
E. x = y or no specific relation can be established

Correct Answer -E. If x = y or no specific relation can be established

First equation –
3x2– 11x + 6 = 0
or, 3x2– 9x – 2x + 6 = 0
or, 3x (x – 3) – 2 (x – 3) = 0
or, (3x – 2) (x – 3) = 0
or, x = 2/3 or 3
Second equation –
2y2– 3y – 9 = 0
or, 2y2– 6y + 3y – 9 = 0
or, 2y (y – 3) + 3 (y – 3) = 0
or, (2y + 3) (y – 3) = 0
or, y = 3 or – 3/2
Here, x can be more than y or less than y so the answer is 5.

9.Based on the equations given choose the option which is true –

I. 6x2-11x+3=0
II. 2y2-7y+6=0
A. x ≤ y
B. x ≥ y
C. x < y
D. x > y
E. x = y or no specific relation can be established

Correct Answer -A. x ≤ y

First equation –
6x2– 11x + 3 = 0
or, 6x2– 9x – 2x + 3 = 0
or, 3x (2x – 3) – 1 (2x – 3) = 0
or, (3x – 1) (2x – 3) = 0
or, x = 1/3 or 3/2
Second equation –
2y2– 7y + 6 = 0
or, 2y2– 4y – 3y + 6 = 0
or, 2y (y – 2) – 3 (y – 2) = 0
or, (2y – 3) (y – 2) = 0
or, y = 3/2 or 2
So here y is either equal to x or more than x.

10.Based on the equations given choose the option which is true –

I. 2x2+5x+2=0
II. 4y2-8y-5=0
A. x ≤ y
B. x ≥ y
C. x < y
D. x > y
E. x = y or no specific relation can be established

Correct Answer – A. If x ≤ y

First equation –
2x2 + 5x + 2 = 0
or, 2x2 + 4x + x + 2 = 0
or, 2x (x + 2) + 1 (x + 2) = 0
or, (2x + 1) (x + 2) = 0
or, x = – ½ or – 2
Second equation –
4y2– 8y – 5 = 0
or, 4y2– 10y + 2y – 5 = 0
or, 2y (2y – 5) + 1 (2y – 5) = 0
or, (2y + 1) (2y – 5) = 0
or, y = – ½ or 5/2
So here y is always more than x or equal to x.