Quadratic Equations

A. 29
B. 27
C. 28
D. 7
Explanation:
Let the roots of the quadratic equation be x and 3x.
Sum of roots = -(-12) = 12
a + 3a = 4a = 12 => a = 3
Product of the roots = 3a2 = 3(3)2 = 27.

A. 9, 10
B. 10, 11
C. 11, 12
D. 12, 13
Explanation:
Let the two consecutive positive integers be x and x + 1
x2 + (x + 1)2 – x(x + 1) = 91
x2 + x – 90 = 0
(x + 10)(x – 9) = 0 => x = -10 or 9.
As x is positive x = 9
Hence the two consecutive positive integers are 9 and 10.

A. 3
B. 4
C. 5
D. 6
Let the roots of the equation 2a and 3a respectively.
2a + 3a = 5a = -(- 5/2) = 5/2 => a = 1/2
Product of the roots: 6a2 = b/2 => b = 12a2
a = 1/2, b = 3.

A. 10, 3
B. -10, 3
C. -20, 3
D. -10, -3
Sum of the roots and the product of the roots are -20 and 3 respectively.

A. x2 + 13x – 140 = 0
B. x2 – 13x + 140 = 0
C. x2 – 13x – 140 = 0
D. x2 + 13x + 140 = 0
Explanation:
Any quadratic equation is of the form
x2 – (sum of the roots)x + (product of the roots) = 0 —- (1)
where x is a real variable. As sum of the roots is 13 and product of the roots is -140, the quadratic equation with roots as 20 and -7 is: x2 – 13x – 140 = 0.

A. rational and unequal
B. complex
C. real and equal
D. irrational and unequal
Explanation:
The discriminant of the quadratic equation is (-12)2 – 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal.

A. 3, -3/2
B. 3/2, -3
C. -3/2, -3
D. 3/2, 3
Explanation:
2×2 + 6x – 3x – 9 = 0
2x(x + 3) – 3(x + 3) = 0
(x + 3)(2x – 3) = 0
=> x = -3 or x = 3/2.

A. -5, 3
B. 3, 5
C. -3, 5
D. -3, -5
Explanation:
x2 + 5x – 3x – 15 = 0
x(x + 5) – 3(x + 5) = 0
(x – 3)(x + 5) = 0
=> x = 3 or x = -5.