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Quadratic Equations Questions with Solutions

This topic isn’t about solving one quadratic equation — it’s about solving two of them (one in x, one in y) and then comparing every possible pair of roots to decide the relationship between x and y. Here’s the concept, four fully worked examples covering each possible relationship, and a few to try yourself.

The concept, quickly

  • Solve equation I for all values of x, and equation II for all values of y — each quadratic usually gives two roots.
  • Compare every x with every y. If x is bigger in every single combination, the answer is "x > y"; if x is always smaller, it’s "x < y".
  • If some combinations give x bigger and others give x equal, the answer is "x ≥ y" (and symmetrically "x ≤ y") — not "x > y", since equality also occurs.
  • If the combinations genuinely conflict — some give x > y, others x < y (or x = y) — no single relationship covers every case, so the answer is "cannot be determined".

Worked examples

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

Answer: x ≥ y

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

Answer: x ≤ y

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

Answer: x > y

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

Answer: x < y

Try these yourself

Solve both equations, compare every pair of roots, then check against the answer.

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

Answer: x = y

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

Answer: Cannot be determined

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

Answer: x < y

Where this comes up

Quadratic Equations is a fixed, dedicated set of questions in the Quantitative Aptitude section of IBPS PO, IBPS Clerk, SBI PO and SBI Clerk, and appears in SSC CGL and SSC CHSL as well.

Practice more Quadratic Equations questions

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