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Quadratic equation questions with answers

Bank papers do not ask you to solve a quadratic. Quadratic equation questions here give you two of them and ask how x compares with y. That is a different skill, and the answer is often "no relation" because the root ranges overlap. Every walkthrough below finds both pairs of roots and then compares the ranges explicitly, which is the step candidates skip.

Download the PDF — 15 questions with answers

Free, ungated, no email. This is the quadratic equation questions with answers PDF — the same 15 questions as below, with every worked solution at the back, ready to print.

How to use this set

The 15 questions below are ordered the way an exam block is rather than easiest-first: two gentle openers, then a run of medium, and a hard one every sixth question — 2 of the 15 sit in the hard band. Every question is a quadratic equations.

Work each question on paper before you open its solution. The walkthrough is the method itself, step for step, so reading it first turns a practice set into a reading exercise and teaches nothing. If you are stuck on the method rather than on one question, the method guide is a better place to start than the next question: Quadratic comparison: the range check.

15 quadratic equation questions with answers

Every question below was generated by the same engine that mints the ComputePrep daily timed drill, and machine-checked before it reached this page — a multiple-choice question validates its own answer against its own options at generation, and ships only with a walkthrough attached. The walkthrough is the engine's own.

Question 1

Solve both equations and state the relation between x and y.

I. x² − 4x + 3 = 0

II. y² − 10y + 24 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: x² − 4x + 3 = 0 → (x − 3)(x − 1) = 0, so x = 1 and 3.
  2. Factorise II: y² − 10y + 24 = 0 → (y − 4)(y − 6) = 0, so y = 4 and 6.
  3. Compare every x against every y — one pair is never enough.
  4. The largest x (3) is still below the smallest y (4), so x < y.

Question 2

Solve both equations and state the relation between x and y.

I. x² − 12x + 36 = 0

II. y² − 5y + 4 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: x² − 12x + 36 = 0 → (x − 6)(x − 6) = 0, so x = 6 (twice).
  2. Factorise II: y² − 5y + 4 = 0 → (y − 4)(y − 1) = 0, so y = 1 and 4.
  3. Compare every x against every y — one pair is never enough.
  4. The smallest x (6) still beats the largest y (4), so x > y.

Question 3

Solve both equations and state the relation between x and y.

I. x² + 13x + 42 = 0

II. y² + 10y + 24 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: x² + 13x + 42 = 0 → (x + 7)(x + 6) = 0, so x = -7 and -6.
  2. Factorise II: y² + 10y + 24 = 0 → (y + 6)(y + 4) = 0, so y = -6 and -4.
  3. Compare every x against every y — one pair is never enough.
  4. The largest x equals the smallest y (-6), and every other x is smaller — so x ≤ y, not x < y.

Question 4

Solve both equations and state the relation between x and y.

I. x² − 8x + 16 = 0

II. y² + 6y − 7 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: x² − 8x + 16 = 0 → (x − 4)(x − 4) = 0, so x = 4 (twice).
  2. Factorise II: y² + 6y − 7 = 0 → (y − 1)(y + 7) = 0, so y = -7 and 1.
  3. Compare every x against every y — one pair is never enough.
  4. The smallest x (4) still beats the largest y (1), so x > y.

Question 5

Solve both equations and state the relation between x and y.

I. x² − 11x + 24 = 0

II. y² − 20y + 96 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: x² − 11x + 24 = 0 → (x − 3)(x − 8) = 0, so x = 3 and 8.
  2. Factorise II: y² − 20y + 96 = 0 → (y − 8)(y − 12) = 0, so y = 8 and 12.
  3. Compare every x against every y — one pair is never enough.
  4. The largest x equals the smallest y (8), and every other x is smaller — so x ≤ y, not x < y.

Question 6

Solve both equations and state the relation between x and y.

I. 2x² − 2x − 4 = 0

II. y² + 3y + 2 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: 2x² − 2x − 4 = 0 → 2(x + 1)(x − 2) = 0, so x = -1 and 2.
  2. Factorise II: y² + 3y + 2 = 0 → (y + 2)(y + 1) = 0, so y = -2 and -1.
  3. Compare every x against every y — one pair is never enough.
  4. The smallest x equals the largest y (-1), and every other x is larger — so x ≥ y, not x > y.

Question 7

Solve both equations and state the relation between x and y.

I. x² − 20x + 100 = 0

II. y² − 11y + 30 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: x² − 20x + 100 = 0 → (x − 10)(x − 10) = 0, so x = 10 (twice).
  2. Factorise II: y² − 11y + 30 = 0 → (y − 6)(y − 5) = 0, so y = 5 and 6.
  3. Compare every x against every y — one pair is never enough.
  4. The smallest x (10) still beats the largest y (6), so x > y.

Question 8

Solve both equations and state the relation between x and y.

I. x² − 9x + 18 = 0

II. y² − 18y + 80 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: x² − 9x + 18 = 0 → (x − 3)(x − 6) = 0, so x = 3 and 6.
  2. Factorise II: y² − 18y + 80 = 0 → (y − 8)(y − 10) = 0, so y = 8 and 10.
  3. Compare every x against every y — one pair is never enough.
  4. The largest x (6) is still below the smallest y (8), so x < y.

Question 9

Solve both equations and state the relation between x and y.

I. x² − 6x + 8 = 0

II. y² − 4 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: x² − 6x + 8 = 0 → (x − 2)(x − 4) = 0, so x = 2 and 4.
  2. Factorise II: y² − 4 = 0 → (y + 2)(y − 2) = 0, so y = -2 and 2.
  3. Compare every x against every y — one pair is never enough.
  4. The smallest x equals the largest y (2), and every other x is larger — so x ≥ y, not x > y.

Question 10

Solve both equations and state the relation between x and y.

I. x² + 20x + 99 = 0

II. y² + 17y + 72 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: x² + 20x + 99 = 0 → (x + 11)(x + 9) = 0, so x = -11 and -9.
  2. Factorise II: y² + 17y + 72 = 0 → (y + 9)(y + 8) = 0, so y = -9 and -8.
  3. Compare every x against every y — one pair is never enough.
  4. The largest x equals the smallest y (-9), and every other x is smaller — so x ≤ y, not x < y.

Question 11

Solve both equations and state the relation between x and y.

I. x² − 5x + 4 = 0

II. y² + 7y + 12 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: x² − 5x + 4 = 0 → (x − 1)(x − 4) = 0, so x = 1 and 4.
  2. Factorise II: y² + 7y + 12 = 0 → (y + 4)(y + 3) = 0, so y = -4 and -3.
  3. Compare every x against every y — one pair is never enough.
  4. The smallest x (1) still beats the largest y (-3), so x > y.

Question 12

Solve both equations and state the relation between x and y.

I. 3x² − 18x + 24 = 0

II. y² + 5y = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: 3x² − 18x + 24 = 0 → 3(x − 2)(x − 4) = 0, so x = 2 and 4.
  2. Factorise II: y² + 5y = 0 → (y − 0)(y + 5) = 0, so y = -5 and 0.
  3. Compare every x against every y — one pair is never enough.
  4. The smallest x (2) still beats the largest y (0), so x > y.

Question 13

Solve both equations and state the relation between x and y.

I. x² − 18x + 80 = 0

II. y² − 14y + 48 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: x² − 18x + 80 = 0 → (x − 8)(x − 10) = 0, so x = 8 and 10.
  2. Factorise II: y² − 14y + 48 = 0 → (y − 6)(y − 8) = 0, so y = 6 and 8.
  3. Compare every x against every y — one pair is never enough.
  4. The smallest x equals the largest y (8), and every other x is larger — so x ≥ y, not x > y.

Question 14

Solve both equations and state the relation between x and y.

I. x² − 8x + 12 = 0

II. y² − 19y + 90 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: x² − 8x + 12 = 0 → (x − 2)(x − 6) = 0, so x = 2 and 6.
  2. Factorise II: y² − 19y + 90 = 0 → (y − 9)(y − 10) = 0, so y = 9 and 10.
  3. Compare every x against every y — one pair is never enough.
  4. The largest x (6) is still below the smallest y (9), so x < y.

Question 15

Solve both equations and state the relation between x and y.

I. x² − 22x + 121 = 0

II. y² − 14y + 45 = 0

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y
  5. x = y or no relation can be established
Show the worked solution
  1. Factorise I: x² − 22x + 121 = 0 → (x − 11)(x − 11) = 0, so x = 11 (twice).
  2. Factorise II: y² − 14y + 45 = 0 → (y − 5)(y − 9) = 0, so y = 5 and 9.
  3. Compare every x against every y — one pair is never enough.
  4. The smallest x (11) still beats the largest y (9), so x > y.
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Where these questions come from

Most free practice for this topic is a scanned upload or a blog quiz, and neither can tell you a question has exactly one answer. That is the one thing this set can promise: it is minted, not typed. The engine behind it serves ComputePrep's daily timed drill, so the questions here are the same shape as the ones a live round would give you — and there is an unlimited supply of them, which is why this page can be regenerated rather than padded.

If you want the method rather than more questions, Quadratic comparison: the range check covers the approach, the traps and a worked table. This page and that one deliberately answer different questions: that one is how do I solve these, this one is give me quadratic equation questions with the answers.

FAQ

When is the answer "no relation can be established"?

Whenever the two root ranges overlap — if x can be larger than one y and smaller than the other, no single relation holds. This option is correct far more often than candidates expect, and skipping the comparison step is what makes it look like a trick.

What is the fastest way to factorise an exam quadratic?

Read the signs first. With x² + bx + c, both roots are negative when b and c are positive, both positive when b is negative and c positive, and of opposite sign when c is negative. That narrows the factor pair before you try a single split.

Do I ever need the quadratic formula in these questions?

Almost never. Exam quadratics are built to factorise with small integer roots. If a split is not appearing within about ten seconds, re-check the sign reading rather than reaching for the discriminant — the formula costs more time than the question is worth.

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