Quadratic Equations(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

QUADRATIC EQUATIONS MCQs

Quadratic Equations, 10th And 12th Grade Quadratic Equations

Total Questions : 102 | Page 8 of 11 pages

x^{2} + y^{2} = 25, x y = 12

,then complete set of x =

{3, 4}
{3, -3}
{3, 4, -3, -4}
{-3, -3}

Discuss Question

Answer: Option C. -> {3, 4, -3, -4}
:
C

x^{2} + y^{2} = 25

and

x y = 12

⇒

x^{2} + (\frac{12}{x})^{2} = 25

⇒

x^{4} + 144 - 25 x^{2} = 0

⇒

(x^{2} - 16) (x^{2} - 9) = 0

⇒

x^{2} = 16

and

x^{2} = 9

⇒x= ±4 and x = ±3

Question 72. If the equation

x^{2} + p x + q = 0

and

x^{2} + q x + p = 0

, have a common root, then p + q + 1 =

0
1
2
-1

Discuss Question

Answer: Option A. -> 0
:
A
Letα is the common root, so

α^{2} + p α + q = 0

and

α^{2} + q α + p = 0

⇒ (p - q)α + (q - p) = 0 ⇒ α=1
Put the value ofα , p + q + 1 = 0

Question 73. If x is real and satisfies x + 2 >

\sqrt{x + 4}

, then

x < -2
x >0
-3 < x < 0
-3 < x < 4

Discuss Question

Answer: Option B. -> x >0
:
B
Given, x + 2 >

\sqrt{x + 4}

⇒

(x + 2)^{2} > (x + 4)

but inequality sign will be remained same if both sides are positive or negative.
R.H.S is positive and so L.H.S will also be positive.
Thus x>-2
⇒

x^{2} + 4 x + 4 > x + 4

⇒

x^{2} + 3 x > 0

⇒

x (x + 3) > 0

⇒

x < - 3 o r x > 0

⇒

x > 0

Question 74. I. x2 - x - 42 = 0, II. y2 - 17y + 72 = 0 to solve both the equations to find the values of x and y?

If x < y
If x > y
If x â‰¤ y
If x â‰¥ y
If x = y or the relationship between x and y cannot be established.

Discuss Question

Answer: 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

Question 75. I. x2 + 9x + 20 = 0, II. y2 + 5y + 6 = 0 to solve both the equations to find the values of x and y?

If x < y
If x > y
If x â‰¤ y
If x â‰¥ y
If x = y or the relationship between x and y cannot be established.

Discuss Question

Answer: 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.

Question 76. I. x2 + 3x - 18 = 0, II. y2 + y - 30 = 0 to solve both the equations to find the values of x and y?

If x < y
If x > y
If x â‰¤ y
If x â‰¥ y
If x = y or the relationship between x and y cannot be established.

Discuss Question

Answer: 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.

Question 77. I. 9a2 + 18a + 5 = 0, II. 2b2 + 13b + 20 = 0 to solve both the equations to find the values of a and b?

If a > b
If a â‰¥ b
If a < b
If a â‰¤ b
If a = b or the relationship between a and b cannot be established.

Discuss Question

Answer: 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.

Question 78. I. x2 + 11x + 30 = 0, II. y2 + 15y + 56 = 0 to solve both the equations to find the values of x and y?

If x < y
If x > y
If x â‰¤ y
If x â‰¥ y
If x = y or the relationship between x and y cannot be established.

Discuss Question

Answer: 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

Question 79. Find the roots of the quadratic equation: 2x2 + 3x - 9 = 0?

3, -3/2
3/2, -3
-3/2, -3
3/2, 3
2/3, -3

Discuss Question

Answer: 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.

Question 80. If the roots of a quadratic equation are 20 and -7, then find the equation?

x2 + 13x - 140 = 0
x2 - 13x + 140 = 0
x2 - 13x - 140 = 0
x2 + 13x + 140 = 0
None of these

Discuss Question

Answer: 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.