Quadratic equations for IBPS PO

Quadratic equation questions give you two equations, one in x and one in y. You solve both, compare every value of x with every value of y, and pick the relation. There is nothing to interpret and no data to read. With practice, each question takes 30 to 40 seconds, which makes a set of these one of the most reliable blocks of marks in the prelims.

The whole skill is three steps done quickly and without slips: factorise, find the roots, compare. This guide covers each step, the shortcut that turns factorising into mental work, and the traps that cost careful candidates their marks.

How the question is asked

Each question gives two equations, labelled I and II, and a fixed set of answer options. The options usually are:

• x > y
• x < y
• x ≥ y
• x ≤ y
• x = y, or the relation cannot be established

Read the last option in your paper carefully. In many papers "x = y" and "cannot be established" share one option, so the two cases lead to the same answer.

Step 1: factorise fast

For x² + bx + c = 0, find two numbers that multiply to c and add to b.

• x² − 7x + 12 = 0 → the numbers are −3 and −4 → (x − 3)(x − 4) = 0 → x = 3 or 4.

For ax² + bx + c = 0, find two numbers that multiply to a × c and add to b, then split the middle term.

• 2x² − 7x + 3 = 0 → a × c = 6 → the numbers are −6 and −1 → 2x² − 6x − x + 3 = 2x(x − 3) − 1(x − 3) = (2x − 1)(x − 3) → x = ½ or 3.

Step 2: the sign-flip shortcut

You do not need to write out the factors. Once you have the two numbers p and q (which multiply to ac and add to b), the roots are −p/a and −q/a. In words: flip the signs of the two numbers and divide each by a.

• 2x² − 7x + 3 = 0: p and q are −6 and −1. Flip: 6 and 1. Divide by 2: 3 and ½.

Why it works: if pq = ac and p + q = b, then (ax + p)(ax + q) = a²x² + a(p + q)x + pq = a²x² + abx + ac = a(ax² + bx + c). So ax² + bx + c = 0 exactly when ax + p = 0 or ax + q = 0, which gives x = −p/a or x = −q/a.

Step 3: know the signs before you solve

Why: the product of the roots is c/a and their sum is −b/a. A positive product means the roots share a sign; a negative sum then means both are negative. A negative product means opposite signs.

This lets you answer some questions without solving at all. If both x roots are positive and both y roots are negative, the answer is x > y, whatever the exact values.

Step 4: compare every pair

Put all the roots on one mental number line and compare each x with each y.

The square-root trap

• x² = 144 means x = +12 or −12. Both roots count.
• x = √144 means x = +12 only. The √ symbol stands for the positive root.

This single difference decides many questions. Compare x = √144 with y² = 144: x = 12, y = ±12, so every x is greater than or equal to every y: x ≥ y.

Similarly, x³ = 343 has one real root, x = 7, while y² = 49 gives y = ±7. Again x ≥ y.

Six worked examples

Example 1: I. x² − 7x + 12 = 0. II. y² − 9y + 20 = 0.

• x = 3, 4. y = 4, 5.
• Every x is at most every y, and 4 = 4. Answer: x ≤ y.

Example 2: I. x² + 5x + 6 = 0. II. y² + 7y + 12 = 0.

• x = −2, −3. y = −3, −4.
• −2 is greater than both y values; −3 equals −3 and is greater than −4. Answer: x ≥ y.

Example 3: I. 2x² − 7x + 3 = 0. II. y² − 4y + 3 = 0.

• x = ½, 3. y = 1, 3.
• ½ is less than 1, but 3 is greater than 1. Answer: relation cannot be established.

Example 4: I. 6x² − 17x + 12 = 0. II. 12y² − 25y + 12 = 0.

• For x: a × c = 72. Two numbers multiplying to 72 and adding to −17: −8 and −9. Flip and divide by 6: x = 8/6 = 4/3 and 9/6 = 3/2.
• For y: a × c = 144. Two numbers multiplying to 144 and adding to −25: −9 and −16. Flip and divide by 12: y = 9/12 = 3/4 and 16/12 = 4/3.
• x = 1.33, 1.5; y = 0.75, 1.33. Every x is at least every y, with 4/3 equal. Answer: x ≥ y.

Example 5: I. 3x² + 11x + 10 = 0. II. 2y² + 7y + 6 = 0.

• Both have positive b and c, so all roots are negative. Solving is still needed.
• For x: a × c = 30; the numbers are 5 and 6. Flip and divide by 3: x = −5/3 ≈ −1.67 and −2.
• For y: a × c = 12; the numbers are 3 and 4. Flip and divide by 2: y = −3/2 = −1.5 and −2.
• −1.67 is less than −1.5 but greater than −2. Answer: relation cannot be established.

Example 6 (a linear pair): I. 3x + 2y = 21. II. 2x + 3y = 19.

• Add them: 5x + 5y = 40, so x + y = 8.
• Subtract II from I: x − y = 2.
• So x = 5 and y = 3. Answer: x > y.
• Check: 15 + 6 = 21 and 10 + 9 = 19.

Common mistakes

• Flipping signs twice, or forgetting to flip them at all. Write the two numbers, flip, divide, and check one root by substituting it.
• Dropping the negative root of x² = k.
• Treating √k as ±. The √ sign gives the positive root only.
• Comparing only the largest roots. Compare every pair, or use the number line.
• Choosing x ≥ y when no roots are equal. If every x is strictly greater, the answer is x > y.

Practice set

1. I. x² − 5x + 6 = 0. II. y² − 3y + 2 = 0.
2. I. x² = 16. II. y² − 8y + 16 = 0.
3. I. x² + 9x + 20 = 0. II. y² + 5y + 6 = 0.
4. I. x² − 1 = 0. II. y² − 2y + 1 = 0.
5. I. 3x² − 10x + 8 = 0. II. y² − 3y + 2 = 0.
6. I. 2x² − 11x + 15 = 0. II. 2y² − 9y + 10 = 0.
7. I. x² + 3x − 10 = 0. II. y² − 6y + 8 = 0.
8. I. x = √196. II. y² = 196.

Answers:

1. x ≥ y. x = 2, 3; y = 1, 2. The only equal pair is 2 and 2.
2. x ≤ y. x = ±4; y = 4 (a repeated root). −4 < 4 and 4 = 4.
3. x < y. x = −4, −5; y = −2, −3. Every x is below every y.
4. x ≤ y. x = ±1; y = 1. −1 < 1 and 1 = 1.
5. Relation cannot be established. a × c = 24; numbers −4 and −6; x = 4/3, 2. y = 1, 2. 4/3 is above 1 but below 2.
6. x ≥ y. For x: a × c = 30; numbers −5 and −6; x = 5/2, 3. For y: a × c = 20; numbers −4 and −5; y = 2, 5/2. Every x is at least every y, with 5/2 equal.
7. x ≤ y. x = −5, 2; y = 2, 4. −5 is below both; 2 = 2 and 2 < 4.
8. x ≥ y. x = 14 only; y = ±14.

What to do next

• Practise ten pairs a day for a week using the sign-flip shortcut, checking one root by substitution each time.
• Then time yourself: a set of five in under three minutes.
• Learn the same comparison logic for quantity comparison, where it reappears in a different form.
• Revise the rest of the fast blocks through the quant plan.