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Test: Quadratic Equations- 2


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10 Questions MCQ Test Quantitative Aptitude (Quant) | Test: Quadratic Equations- 2

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Test: Quadratic Equations- 2 - Question 1

Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 8x + 4 = 0?

Detailed Solution for Test: Quadratic Equations- 2 - Question 1

Explanation:

a/b + b/a = (a2 + b2)/ab = (a2 + b2 + a + b)/ab 
= [(a + b)2 - 2ab]/ab
a + b = -8/1 = -8
ab = 4/1 = 4
Hence a/b + b/a = [(-8)2 - 2(4)]/4 = 56/4 = 14.

Test: Quadratic Equations- 2 - Question 2

Detailed Solution for Test: Quadratic Equations- 2 - Question 2

Test: Quadratic Equations- 2 - Question 3

Find the quadratic equations whose roots are the reciprocals of the roots of 2x2 + 5x + 3 = 0?

Detailed Solution for Test: Quadratic Equations- 2 - Question 3

Explanation:

The quadratic equation whose roots are reciprocal of 2x2 + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)2 + 5(1/x) + 3 = 0
=> 3x2 + 5x + 2 = 0

Test: Quadratic Equations- 2 - Question 4

A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?

Detailed Solution for Test: Quadratic Equations- 2 - Question 4

Explanation:

Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x 
(x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy
=>5(300/x) - 10x - 50 = 0 => -150 + x2 + 5x = 0
multiplying both sides by -1/10x
=> x2 + 15x - 10x - 150 = 0
=> x(x + 15) - 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10.

Test: Quadratic Equations- 2 - Question 5

I. a2 - 7a + 12 = 0,
II. b2 - 3b + 2 = 0 to solve both the equations to find the values of a and b?

Detailed Solution for Test: Quadratic Equations- 2 - Question 5

Explanation:

I.(a - 3)(a - 4) = 0
=> a = 3, 4


II. (b - 2)(b - 1) = 0
=> b = 1, 2
=> a > b

Test: Quadratic Equations- 2 - Question 6

I. a2 - 9a + 20 = 0,
II. 2b2 - 5b - 12 = 0 to solve both the equations to find the values of a and b?

Detailed Solution for Test: Quadratic Equations- 2 - Question 6

Explanation:

I. (a - 5)(a - 4) = 0
=> a = 5, 4
II. (2b + 3)(b - 4) = 0
=> b = 4, -3/2 => a ≥ b

Test: Quadratic Equations- 2 - Question 7

 I. a2 + 11a + 30 = 0,
II. b2 + 6b + 5 = 0 to solve both the equations to find the values of a and b?

Detailed Solution for Test: Quadratic Equations- 2 - Question 7

Explanation:

I. (a + 6)(a + 5) = 0
=> a = -6, -5
II. (b + 5)(b + 1) = 0
=> b = -5, -1 => a ≤ b

Test: Quadratic Equations- 2 - Question 8

I. a2 + 8a + 16 = 0,
II. b2 - 4b + 3 = 0 to solve both the equations to find the values of a and b?

Detailed Solution for Test: Quadratic Equations- 2 - Question 8

Explanation:

I. (a + 4)2 = 0 => a = -4
II.(b - 3)(b - 1) = 0
=> b = 1, 3 => a < b

Test: Quadratic Equations- 2 - Question 9

I. a2 - 2a - 8 = 0,
II. b2 = 9 to solve both the equations to find the values of a and b?

Detailed Solution for Test: Quadratic Equations- 2 - Question 9

Explanation:

I. (a - 4)(a + 2) = 0
=> a = 4, -2
II. b2 = 9
=> b = ± 3
-2 < 3, -2 > -3, 4 > 3, 4 > -3,
No relation can be established between a and b.

Test: Quadratic Equations- 2 - Question 10

I. x2 + 5x + 6 = 0,
II. y2 + 9y +14 = 0 to solve both the equations to find the values of x and y?

Detailed Solution for Test: Quadratic Equations- 2 - Question 10

I. x2 + 3x + 2x + 6 = 0
=> (x + 3)(x + 2) = 0 => x = -3 or -2
II. y2 + 7y + 2y + 14 = 0
=> (y + 7)(y + 2) = 0 => y = -7 or -2
No relationship can be established between x and y.

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