Quadratic Equation Questions Quiz-3

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

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

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.

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

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

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

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.

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.

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.

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.