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Quantitative Aptitude

QUADRATIC EQUATIONS MCQs

Total Questions : 30

Page 1 of 3 pages
Question 1. I. x2 - x - 42 = 0, II. y2 - 17y + 72 = 0 to solve both the equations to find the values of x and y?
  1.    If x < y
  2.    If x > y
  3.    If x ≤ y
  4.    If x ≥ y
  5.    If x = y or the relationship between x and y cannot be established.
 Discuss Question
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

Question 2. I. x2 + 9x + 20 = 0, II. y2 + 5y + 6 = 0 to solve both the equations to find the values of x and y?
  1.    If x < y
  2.    If x > y
  3.    If x ≤ y
  4.    If x ≥ y
  5.    If x = y or the relationship between x and y cannot be established.
 Discuss Question
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.

Question 3. I. x2 + 3x - 18 = 0, II. y2 + y - 30 = 0 to solve both the equations to find the values of x and y?
  1.    If x < y
  2.    If x > y
  3.    If x ≤ y
  4.    If x ≥ y
  5.    If x = y or the relationship between x and y cannot be established.
 Discuss Question
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.

Question 4. I. 9a2 + 18a + 5 = 0, II. 2b2 + 13b + 20 = 0 to solve both the equations to find the values of a and b?
  1.    If a > b
  2.    If a ≥ b
  3.    If a < b
  4.    If a ≤ b
  5.    If a = b or the relationship between a and b cannot be established.
 Discuss Question
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.

Question 5. I. x2 + 11x + 30 = 0, II. y2 + 15y + 56 = 0 to solve both the equations to find the values of x and y?
  1.    If x < y
  2.    If x > y
  3.    If x ≤ y
  4.    If x ≥ y
  5.    If x = y or the relationship between x and y cannot be established.
 Discuss Question
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

Question 6. Find the roots of the quadratic equation: 2x2 + 3x - 9 = 0?
  1.    3, -3/2
  2.    3/2, -3
  3.    -3/2, -3
  4.    3/2, 3
  5.    2/3, -3
 Discuss Question
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.

Question 7. If the roots of a quadratic equation are 20 and -7, then find the equation?
  1.    x2 + 13x - 140 = 0
  2.    x2 - 13x + 140 = 0
  3.    x2 - 13x - 140 = 0
  4.    x2 + 13x + 140 = 0
  5.    None of these
 Discuss Question
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.

Question 8. The roots of the equation 3x2 - 12x + 10 = 0 are?
  1.    rational and unequal
  2.    complex
  3.    real and equal
  4.    irrational and unequal
  5.    rational and equal
 Discuss Question
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.

Question 9. The sum and the product of the roots of the quadratic equation x2 + 20x + 3 = 0 are?
  1.    10, 3
  2.    -10, 3
  3.    20, -3
  4.    -10, -3
  5.    None of these
 Discuss Question
Answer is Option E. -> None of these

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

Question 10. Find the roots of the quadratic equation: x2 + 2x - 15 = 0?
  1.    -5, 3
  2.    3, 5
  3.    -3, 5
  4.    -3, -5
  5.    5, 2
 Discuss Question
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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