Quadratic Equation Questions Quiz-5

B. x < y

x² + 5x + 6 =0
x²+ 2x + 3x +6 =0
(x + 2)(x + 3)
x = –2, –3
(y²/2) – 4.5y + 10 = 0
y²– 9y + 20 = 0

=> y²– 4y – 5y + 4*5 = 0
(y – 4)(y – 5)
y = 4, 5

x<y

B. x < y

y²– 13y + 40 = 0
y²– 8y – 5y + 5*8 = 0
y(y – 8) – 5(y – 8) = 0
(y – 5)(y – 8) = 0
y = 5, 8
6x²– 7x + 2 = 0
6x²– 4x – 3x + 2 = 0
2x(3x – 2) – (3x – 2) = 0
(2x – 1)(3x – 2) = 0
x = 1/2 or 2/3
x<y

C. if x = y, or relation cannot be established between x and y

15x²+ 57x – 12 = 0
15x²+ 60x – 3x – 12 = 0
15x(x + 4) – 3(x + 4) = 0
(15x – 3)(x + 4) = 0
3(5x – 1)(x + 4) = 0
x = 1/5, -4
6y²+ 25y – 9 = 0
6y²+ 27y – 2y – 9 = 0
3y(2y + 9) – (2y + 9) = 0
(3y – 1)(2y + 9)
y = 1/3 or -9/2
Therefore, relationship between x and y can’t be determined.

C. if x = y, or relation cannot be established between x and y

4x⁴– 5x²+ 1 = 0
4x⁴– 4x²– x²+ 1 = 0
4x²(x²– 1) – 1(x²– 1) = 0
(4x²– 1) (x²– 1) = 0
x = 1, -1, 1/2, -1/2
y2– y – 2 = 0
y2+ y – 2y – 2 = 0
y(y + 1) – 2 (y + 1) = 0
(y – 2) (y + 1) = 0
y = 2, -1
Therefore, relationship between x and y can’t be determined.

C. if x = y, or relation cannot be established between x and y

14y²+ 75y – 11 = 0
14y²+ 77y – 2y – 11 = 0
7y(2y + 11) – (2y + 11) = 0
(7y-1) (2y + 11)
y = 1/7, -11/2
6x²+ 7x + 2 = 0
6x²+ 3x + 4x + 2 = 0
3x(2x + 1) + 2(2x + 1) = 0
(3x + 2)(2x + 1) = 0
x = -2/3 or -1/2
Therefore, relationship between x and y can’t be determined.

A. if x > y

2x²– 37x + 143 = 0
=> 2x²– 26x – 11x + 143 = 0
=> 2x(x-13) – 11(x-13) = 0
=> (x-13)(2x-11) = 0
=> x = 13, 11/2
3y²+ 31y – 22 = 0
3y²+ 33y – 2y – 22 = 0
=> 3y(y+11) – 2(y+11) = 0
=> (y+11)(3y-2)
=> y = -11, 2/3
Therefore, x > y

B. if x < y

6×2 + 31x + 35 = 0
=> 6x² + 21x + 10x + 35 = 0
=> 3x(2x+7) + 5(2x+7) = 0
=> (2x+7)(3x+5) = 0
x = -7/2, -5/3
4y²– 8y – 21 = 0
=> 4y²– 8y – 21 = 0
=> 4y² + 6y – 14y – 21 = 0
=> (2y+3)(2y-7) = 0
=> y = -3/2, 7/2
Therefore, x < y

C. if x = y, or relation cannot be established between x and y

2x²– 11x – 21 = 0
=> 2x²– 14x + 3x – 21 = 0
=> 2x(x-7) + 3(x-7) = 0
=> (2x+3)(x-7) = 0
=> x = -3/2, 7
y²– 3y – 28 = 0
=> y²– 7y + 4y – 28 = 0
=> y(y-7) + 4(y-7) = 0
=> (y+4)(y-7) = 0
=> y = -4, 7
Therefore, no relation can be established between x and y.

C. if x = y, or relation cannot be established between x and y

35x²– 74x + 35 = 0
=> 35x²– 49x – 25x + 35 = 0
=> 7x(5x-7) – 5(5x-7) = 0
=> (7x-5)(5x-7) = 0
=> y = 5/7, 7/5
y²– y – 6 = 0
=> y²+ 2y – 3y – 6 = 0
=> y(y+2) – 3(y-2) = 0
=> (y+2)(y-3) = 0
=> y = – 2, 3
Therefore, relationship between x and y can’t be established

C. if x = y, or relation cannot be established between x and y

x²+ 6x – 55 = 0
=> x²+ 11x – 5x – 55 = 0
=> x(x+11) – 5(x+11) = 0
=> (x+11)(x-5) = 0
=>x = -11, 5
3y²– 14y – 5 = 0
=>3y²+ y -15y – 5 = 0
=> y(3y+1) – 5(3y+1) = 0
=> (3y+1)(y-5) = 0
=> y = -1/3, 5
Therefore, no relationship between x and y can be established.