Quantitative Aptitude

Quadratic Equations: Practice Questions with Answers

14 practice questions on Quadratic Equations, with answers; most also have an explanation or a worked solution. Try each one before you open its answer.

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1. Q.1

   I. x² − 8x + 15 = 0 II. y² − 2y − 15 = 0. Find the relationship between x and y.

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   Answer: Cannot be determined

   Solution:

   1. I: x² − 8x + 15 = (x − 3)(x − 5) = 0, so x = 3 or 5.
   2. II: y² − 2y − 15 = (y − 5)(y + 3) = 0, so y = 5 or −3. The negative product of roots (−15) for equation II was a hint that one root would be negative.
   3. Comparing every pair: (3,5)→x<y, (3,−3)→x>y, (5,5)→x=y, (5,−3)→x>y. The results genuinely conflict (both > and < occur), so the relationship cannot be determined.

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2. Q.2

   I. x² − 7x + 12 = 0 II. y² − 3y + 2 = 0. Find the relationship between x and y.

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   Answer: x > y

   Solution:

   1. I: x² − 7x + 12 = (x − 3)(x − 4) = 0, so x = 3 or 4.
   2. II: y² − 3y + 2 = (y − 1)(y − 2) = 0, so y = 1 or 2.
   3. Every value of x (3 or 4) is bigger than every value of y (1 or 2) — all four combinations give x>y with no equality possible.

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3. Q.3

   I. x² − 9x + 20 = 0 II. y² − 11y + 30 = 0. Find the relationship between x and y.

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   Answer: x ≤ y

   Solution:

   1. I: x² − 9x + 20 = (x − 4)(x − 5) = 0, so x = 4 or 5.
   2. II: y² − 11y + 30 = (y − 5)(y − 6) = 0, so y = 5 or 6.
   3. Comparing every pair: (4,5)→x<y, (4,6)→x<y, (5,5)→x=y, (5,6)→x<y. x is never bigger than y, but can equal it — so the relationship is x ≤ y.

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4. Q.4

   Solve x² − 4x − 1 = 0 using the quadratic formula.

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   Answer: x = 2 + √5 or 2 − √5 (≈ 4.24 or ≈ −0.24)

   Solution:

   1. Here a=1, b=−4, c=−1. Discriminant D = b² − 4ac = 16 − 4(1)(−1) = 16 + 4 = 20.
   2. x = [−b ± √D] / 2a = [4 ± √20] / 2 = [4 ± 2√5] / 2 = 2 ± √5.
   3. So x = 2 + √5 ≈ 4.24, or x = 2 − √5 ≈ −0.24.

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5. Q.5

   I. x² − 3x + 2 = 0 II. y² − 7y + 12 = 0. Find the relationship between x and y.

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   Answer: x < y

   Solution:

   1. I: x² − 3x + 2 = (x − 1)(x − 2) = 0, so x = 1 or 2.
   2. II: y² − 7y + 12 = (y − 3)(y − 4) = 0, so y = 3 or 4.
   3. Every value of x (1 or 2) is smaller than every value of y (3 or 4) — all four combinations give x<y with no equality possible.

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6. Q.6

   Determine the nature of the roots of x² + 2x + 5 = 0, without fully solving it.

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   Answer: No real roots (D = −16 < 0)

   Solution:

   1. a=1, b=2, c=5. Discriminant D = b² − 4ac = 4 − 20 = −16.
   2. Since D < 0, the equation has no real roots (its roots are complex).

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7. Q.7

   I. x² − 11x + 30 = 0 II. y² − 9y + 20 = 0. Find the relationship between x and y.

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   Answer: x ≥ y

   Solution:

   1. I: x² − 11x + 30 = (x − 5)(x − 6) = 0, so x = 5 or 6.
   2. II: y² − 9y + 20 = (y − 4)(y − 5) = 0, so y = 4 or 5.
   3. Comparing every pair: (5,4)→x>y, (5,5)→x=y, (6,4)→x>y, (6,5)→x>y. x is never smaller than y, but can equal it — so the relationship is x ≥ y.

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8. Q.8

   Determine the nature of the roots of 2x² − 3x + 5 = 0 without solving.

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   Answer: No real roots. D = 9−40 = −31 < 0.

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9. Q.9

   Determine the nature of the roots of x² − 6x + 9 = 0 without solving.

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   Answer: Real and equal roots. D = 36−36 = 0.

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10. Q.10

    I. x² − 5x + 6 = 0 II. y² − 4y + 3 = 0. Find the relationship between x and y.

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    Answer: Cannot be determined

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11. Q.11

    I. x² − 4x + 3 = 0 II. y² − 11y + 30 = 0. Find the relationship between x and y.

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    Answer: x < y

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12. Q.12

    Solve x² − 6x − 2 = 0 using the quadratic formula.

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    Answer: x = 3 ± √11. D = 36+8 = 44; x = (6±√44)/2.

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13. Q.13

    I. x² − x − 12 = 0 II. y² − 7y + 10 = 0. Find the relationship between x and y.

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    Answer: Cannot be determined. x = 4 or −3; y = 2 or 5; pairs give both x>y and x<y.

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14. Q.14

    I. x² − 6x + 9 = 0 II. y² − 6y + 9 = 0. Find the relationship between x and y.

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    Answer: x = y

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