Quadratic equations and inequalities for SBI PO

In these questions you are given two equations, one in x and one in y. You solve both, then decide how x and y compare. They appear in bank PO prelims, often as a small block, and once factorising is automatic each one takes 20 to 30 seconds.

The maths is school algebra. The marks are lost in two places: slow factorising, and careless comparison of the roots. This guide fixes both.

The answer options

The options usually look like this:

• x > y
• x < y
• x ≥ y
• x ≤ y
• x = y, or no relation can be established

The relation must hold for every pair of values. Each quadratic usually gives two roots, so you are comparing two values of x with two values of y: four comparisons in all.

Factorising

When the x² term has coefficient 1 (x² + bx + c = 0): find two numbers that multiply to c and add to b.

• x² − 11x + 30 = 0: −5 and −6 multiply to 30 and add to −11. So (x − 5)(x − 6) = 0 and x = 5 or 6.

When there is a leading coefficient (ax² + bx + c = 0): find two numbers that multiply to a × c and add to b, then split the middle term.

• 2x² − 7x + 6 = 0: a × c = 12, and −3 and −4 add to −7.
• 2x² − 3x − 4x + 6 = x(2x − 3) − 2(2x − 3) = (x − 2)(2x − 3).
• Roots: 2 and 1.5.

The quick root trick

You do not need to write out the factorised form. Once you have the two numbers p and q (product a × c, sum b), the roots are −p/a and −q/a.

For 6x² − 13x + 6 = 0: a × c = 36, and −4 and −9 add to −13. Change the signs and divide by a = 6: the roots are 4/6 = 2/3 and 9/6 = 1.5.

Why it works: if p + q = b and pq = ac, then a(ax² + bx + c) = (ax + p)(ax + q). Setting each bracket to zero gives x = −p/a and x = −q/a.

The sign rule

When a is positive, the signs of b and c tell you the signs of the roots before you solve anything.

This follows from sum of roots = −b/a and product = c/a. A positive product means same signs; the sum then shows which sign.

The rule can settle a question instantly. If both x-roots are positive and both y-roots are negative, the answer is x > y without calculating anything.

Square and cube cases

• x² = 49 gives x = 7 or −7. Both signs count.
• x³ = 343 gives only x = 7. A real cube root has one value.
• y = √49 means y = 7 only. The √ symbol denotes the positive root.

Comparing the roots

Put all four values on a mental number line, then read off the relation.

Worked examples

Example 1. I. x² − 11x + 30 = 0 · II. y² − 7y + 12 = 0

• x = 5, 6. y = 3, 4.
• Every x is larger than every y: x > y.

Example 2. I. 6x² − 13x + 6 = 0 · II. 2y² − 9y + 10 = 0

• x: product 36, sum −13 → −4 and −9 → x = 2/3 ≈ 0.67 and 1.5.
• y: product 20, sum −9 → −4 and −5 → y = 4/2 = 2 and 5/2 = 2.5.
• Largest x (1.5) is below smallest y (2): x < y.

Example 3. I. x² = 49 · II. y² − 14y + 49 = 0

• x = 7 or −7. y = 7 (a repeated root, since the expression is (y − 7)²).
• −7 < 7 and 7 = 7: x ≤ y.

Example 4. I. x² + 3x − 10 = 0 · II. y² − y − 2 = 0

• x: (x + 5)(x − 2) → x = −5, 2. y: (y − 2)(y + 1) → y = 2, −1.
• x = −5 is below both y values, but x = 2 is above y = −1: no relation.

Example 5. I. 3x² + 11x + 10 = 0 · II. 2y² + 11y + 15 = 0

• Sign rule: b and c positive in both, so all four roots are negative.
• x: product 30, sum 11 → 5 and 6 → x = −5/3 ≈ −1.67 and −6/3 = −2.
• y: product 30, sum 11 → 5 and 6 → y = −5/2 = −2.5 and −6/2 = −3.
• Smallest x (−2) is above largest y (−2.5): x > y.

Example 6 (linear pair). I. 3x + 2y = 16 · II. x + y = 6

• From II, y = 6 − x. Substitute: 3x + 12 − 2x = 16, so x = 4 and y = 2.
• Only one value each: x > y.

Practice set

1. I. x² − 13x + 42 = 0 · II. y² − 11y + 30 = 0
2. I. 3x² − 10x + 8 = 0 · II. y² − 6y + 9 = 0
3. I. x² + 9x + 18 = 0 · II. y² + 5y + 4 = 0
4. I. 2x² − 11x + 15 = 0 · II. 3y² − 14y + 15 = 0
5. I. x² − 16 = 0 · II. y² − 9y + 20 = 0
6. I. x³ = 343 · II. y² = 49
7. I. 4x² + 7x + 3 = 0 · II. 3y² + 7y + 4 = 0
8. I. 2x + 3y = 14 · II. 4x − y = 14

Answers:

1. x ≥ y. x = 6, 7; y = 5, 6. The only tie is 6 = 6; otherwise x is larger.
2. x < y. x = 4/3, 2 (product 24, sum −10 → −4, −6, divided by 3); y = 3 (repeated). Both x values are below 3.
3. No relation. x = −3, −6; y = −1, −4. x = −3 is below −1 but above −4.
4. No relation. x: product 30, sum −11 → x = 5/2 = 2.5 and 6/2 = 3. y: product 45, sum −14 → −5, −9 → y = 5/3 ≈ 1.67 and 9/3 = 3. x = 2.5 is above 1.67 but below 3.
5. x ≤ y. x = 4 or −4; y = 4, 5. −4 is below both, 4 equals 4 and is below 5.
6. x ≥ y. x = 7 only; y = 7 or −7.
7. x ≥ y. x: product 12, sum 7 → 3, 4 → x = −3/4 = −0.75 and −4/4 = −1. y: product 12, sum 7 → y = −3/3 = −1 and −4/3 ≈ −1.33. The only tie is −1 = −1; otherwise x is larger.
8. x > y. From II, y = 4x − 14. Then 2x + 12x − 42 = 14, so x = 4 and y = 2.

What to do next

• Practise the quick root trick on 20 quadratics with a leading coefficient until it is automatic.
• Time yourself on blocks of five comparisons; aim for under two and a half minutes.
• Use the same root-comparison logic in the mains: read quantity comparison.
• Keep your prelims quick types together with number series and simplification.