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# Elitmus Equations and Inequalities Questions And Answers

1. Find the roots of the quadratic equation: x2 + 2x – 15 = 0?
A. -5, 3
B. 3, 5
C. -3, 5
D. -3, -5

Explaination :

x2 + 5x – 3x – 15 = 0
x(x + 5) – 3(x + 5) = 0
(x – 3)(x + 5) = 0
=> x = 3 or x = -5.

2. Find the roots of the quadratic equation: 2×2 + 3x – 9 = 0?
A. 3, -3/2
B. 3/2, -3
C. -3/2, -3
D. 3/2, 3

Explaination :

2x2 + 6x – 3x – 9 = 0
2x(x + 3) – 3(x + 3) = 0
(x + 3)(2x – 3) = 0
=> x = -3 or x = 3/2.

3. The roots of the equation 3×2 – 12x + 10 = 0 are?
A. rational and unequal
B. complex
C. real and equal
D. irrational and unequal

Explaination :
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.

4. The sum and the product of the roots of the quadratic equation x2 + 20x + 3 = 0 are?
A. 10, 3
B. -10, 3
C. 20, -3
D. None of these

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

5. If the roots of the equation 2×2 – 5x + b = 0 are in the ratio of 2:3, then find the value of b?
A. 3
B. 4
C. 5
D. 6

Explaination :

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.

6. The sum of the square of the three consecutive even natural numbers is 1460. Find the numbers?
A. 18, 20, 22
B. 20, 22, 24
C. 22, 24, 26
D. 24, 26, 28

Explaination :

Three consecutive even natural numbers be 2x – 2, 2x and 2x + 2.
(2x – 2)2 + (2x)2 + (2x + 2)2 = 1460
4x2 – 8x + 4 + 4x2 + 8x + 4 = 1460
12x2 = 1452 => x2 = 121 => x = ± 11
As the numbers are positive, 2x > 0. Hence x > 0. Hence x = 11.
Required numbers are 20, 22, 24.

7. Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 8x + 4 = 0?
A. 15
B. 14
C. 24
D. 26

Explaination :

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.

8. 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?
A. 10
B. 8
C. 15
D. 7.50

Explaination :

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.

9.  I. a2 + 11a + 30 = 0,9.  I. a2 + 11a + 30 = 0,
II. b2 + 6b + 5 = 0 to solve both the equations to find the values of a and b?
A. If a < b
B. If a ≤ b
C. If the relationship between a and b cannot be established
D. If a > b

Explaination :

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

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

A. If x < y

B. If x > y

C. If x ≤ y

D. If x ≥ y

Explaination :

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

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