### Quantitative Aptitude

## QUADRATIC EQUATIONS MCQs

### Total Questions : 30

**Page 1 of 3 pages**

**Answer is Option A. ->**If x < y

I. x2 - 7x + 6x - 42 = 0

=> (x - 7)(x + 6) = 0 => x = 7, -6

II. y2 - 8y - 9y + 72 = 0

=> (y - 8)(y - 9) = 0 => y = 8, 9

=> x < y

**Answer is Option A. ->**If x < y

I. x2 + 4x + 5x + 20 = 0

=>(x + 4)(x + 5) = 0 => x = -4, -5

II. y2 + 3y + 2y + 6 = 0

=>(y + 3)(y + 2) = 0 => y = -3, -2

= x < y.

**Answer is Option E. ->**If x = y or the relationship between x and y cannot be established.

I. x2 + 6x - 3x - 18 = 0

=>(x + 6)(x - 3) = 0 => x = -6, 3

II. y2 + 6y - 5y - 30 = 0

=>(y + 6)(y - 5) = 0 => y = -6, 5

No relationship can be established between x and y.

**Answer is Option A. ->**If a > b

I. 9a2 + 3a + 15a + 5 = 0

=>(3a + 5)(3a + 1) = 0 => a = -5/3, -1/3

II. 2b2 + 8b + 5b + 20 = 0

=>(2b + 5)(b + 4) = 0 => b = -5/2, -4

a is always more than b.

a > b.

**Answer is Option B. ->**If x > y

I. x2 + 6x + 5x + 30 = 0

=>(x + 6)(x + 5) = 0 => x = -6, -5

II. y2 + 8y + 7y + 56 = 0

=>(y + 8)(y + 7) = 0 => y = -8, -7

=> x > y

**Answer is Option B. ->**3/2, -3

2x2 + 6x - 3x - 9 = 0

2x(x + 3) - 3(x + 3) = 0

(x + 3)(2x - 3) = 0

=> x = -3 or x = 3/2.

**Answer is Option C. ->**x2 - 13x - 140 = 0

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.

**Answer is Option D. ->**irrational and unequal

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.

**Answer is Option E. ->**None of these

Sum of the roots and the product of the roots are -20 and 3 respectively.

**Answer is Option A. ->**-5, 3

x2 + 5x - 3x - 15 = 0

x(x + 5) - 3(x + 5) = 0

(x - 3)(x + 5) = 0

=> x = 3 or x = -5.

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