A quadratic equation is any equation of the form ax² + bx + c = 0, where a ≠ 0, and a, b, c are real numbers. The word "quadratic" comes from "quadratus" — square — because the highest power of the unknown is 2.
Here is the key mental model: a quadratic equation has at most two solutions, called roots (or zeros). These roots are the x-values where the parabola y = ax² + bx + c crosses the x-axis. When the parabola touches the axis at exactly one point, the roots are equal. When it never crosses, the roots are complex (no real solutions).
Think of it this way. You have a balance scale. The equation x² - 5x + 6 = 0 is asking: "What two values of x keep the scale at zero?" The answer is x = 2 and x = 3, because both satisfy the equation. These two roots are not random — they have a structured relationship with the coefficients a, b, c through what are called Vieta's formulas.
Vieta's formulas are the backbone of IBPS PO quadratic questions. For ax² + bx + c = 0 with roots α and β:
α + β = -b/aαβ = c/aFor a monic equation (where a = 1), this simplifies beautifully: sum of roots = -b, product of roots = c. Most IBPS PO questions hand you a monic equation, so you will use this simplified form constantly.
The analogy worth burning in: b and c in x² + bx + c are the "fingerprints" of the roots. The sum left a fingerprint at -b, and the product left one at c. Every manipulation question — finding α² + β², forming new equations, computing α³ + β³ — is just forensics on those two fingerprints.
Given ax² + bx + c = 0:
Method 1: Factorization (fastest for IBPS PO)
Split the middle term bx into two terms whose coefficients multiply to ac and add to b. For x² - 5x + 6 = 0: you need two numbers that multiply to 6 and add to -5. Those are -2 and -3. So: x² - 2x - 3x + 6 = x(x - 2) - 3(x - 2) = (x - 2)(x - 3) = 0, giving x = 2 or x = 3.
Method 2: Quadratic Formula (last resort)
x = (-b ± √(b² - 4ac)) / 2a
Use this only when factorization fails — typically when roots are irrational (involving surds). For IBPS PO, clean integer roots dominate, so factorization will handle 80% of questions.
Method 3: Sum-Product direct read (fastest of all, when applicable)
When you see x² - Sx + P = 0 with a monic equation, you can sometimes spot the roots by asking: "what two numbers add to S and multiply to P?" For x² - 8x + 15 = 0, you want numbers adding to 8 and multiplying to 15 — that's 3 and 5. Done. No paper needed.
The discriminant D = b² - 4ac tells you what kind of roots to expect before you solve.
| D value | Nature of roots | |---|---| | D > 0 | Two distinct real roots | | D = 0 | Two equal real roots (repeated root) | | D < 0 | No real roots (complex conjugates) |
For equal roots: set D = 0, which gives you b² = 4ac. This is a direct formula for "find k such that equal roots exist" questions.
Once you extract α + β and αβ from the equation, you can compute almost anything using algebraic identities:
α² + β² = (α + β)² - 2αβα³ + β³ = (α + β)³ - 3αβ(α + β)(α - β)² = (α + β)² - 4αβα/β + β/α = (α² + β²)/(αβ) = [(α + β)² - 2αβ] / αβLook — every one of these expressions reduces to functions of just (α + β) and αβ. You never need to find the individual roots to answer these questions. That is the leverage point.
The reverse of Vieta: if you know the two roots r₁ and r₂, the equation is:
x² - (r₁ + r₂)x + r₁r₂ = 0
Shorthand: x² - (sum)x + (product) = 0.
When roots involve surds like 2 + √3 and 2 - √3, the sum is 4 and the product is (2)² - (√3)² = 4 - 3 = 1, giving x² - 4x + 1 = 0. The difference of squares identity wipes out the surd cleanly.
IBPS PO loves the "form a new equation whose roots are [transformation of original roots]" pattern. The key transformations:
Shift by k: If original roots are α, β, new roots are α + k, β + k.
(α + β) + 2kαβ + k(α + β) + k²x → (x - k) in the original equation.Scale by k: New roots are kα, kβ.
k(α + β), New product = k²αβx → x/k in the original equation.Reciprocals: New roots are 1/α, 1/β.
ax² + bx + c = 0 with cx² + bx + a = 0.The substitution shortcuts let you write the new equation in one step without computing the new sum and product separately.
Many IBPS PO sets give you two separate quadratics — one in x, one in y — and ask you to compare. The comparison types are: x > y, x < y, x ≥ y, x ≤ y, or cannot be determined.
Process:
This is mechanical — no special formula, just accurate factorization followed by systematic comparison.
Before picking up the pen, scan a monic quadratic for integer roots by asking: "two numbers adding to |b| and multiplying to |c|?" For x² - 9x + 20 = 0, scan pairs multiplying to 20: (1,20), (2,10), (4,5). Which adds to 9? That's (4,5). Roots are 4 and 5. Standard method using middle-term split: 6 steps. This scan: 1 step, under 5 seconds.
When roots are conjugate surds (a + √b) and (a - √b), their product is always a² - b — no expansion needed. For roots 3 + √5 and 3 - √5: product = 9 - 5 = 4. Sum = 6. Equation: x² - 6x + 4 = 0. Expanding (3 + √5)(3 - √5) the long way takes 4 sub-steps. Difference of squares: 1 sub-step.
For "find k for equal roots" questions on x² + kx + c = 0, the condition D = 0 gives k² = 4c, so k = ±2√c. For x² + kx + 9 = 0: k = ±2√9 = ±6. No need to write out the full discriminant formula. Standard discriminant expansion: 4 steps. Direct plug: 2 steps.
Memorize this chain: you always have just two facts — S = α + β and P = αβ. Then α³ + β³ = S³ - 3PS. For α + β = 6, αβ = 8: 6³ - 3(8)(6) = 216 - 144 = 72. If you try to find α and β individually first (roots of x² - 6x + 8 = 0 are 2 and 4), then cube each and add: that's 5 steps. The identity chain: 2 steps.
If the new roots are reciprocals of the original roots, just swap the leading coefficient and the constant. Original: 3x² - 7x + 2 = 0. Reciprocal roots equation: 2x² - 7x + 3 = 0. Done. Deriving via 1/α + 1/β = (α + β)/(αβ) and (1/α)(1/β) = 1/(αβ) then forming the equation: 5 steps. Coefficient swap: 1 step.
Step 1 — Classify the question type:
x² - (new sum)x + (new product) = 0.Step 2 — Never compute what Vieta gives for free. If the question asks for S or P directly, read it off the coefficients. Do not find the roots.
Step 3 — Check sign discipline. The single most common error is sign flip in α + β = -b/a. For x² - 7x + 10 = 0, b = -7, so α + β = -(-7)/1 = 7. If you misread as -7, every downstream calculation is wrong.
Why this question: Tests whether you use the product-of-roots shortcut when one root is known, instead of substituting back and solving a linear equation for the parameter.
Solving path: One root is 3. Product of roots = c/a = 12/1 = 12. Other root = 12/3 = 4. Sum of roots = 3 + 4 = 7 = p. Answer: other root is 4, p = 7.
Why this question: Tests the α/β + β/α pattern — a two-layer application of Vieta that catches students who try to find individual roots.
Solving path: Read off p + q = 8, pq = 15. Target: (p² + q²)/(pq). Compute numerator: (p + q)² - 2pq = 64 - 30 = 34. Divide by pq: 34/15. No need to find p and q individually.
Why this question: The "shift roots by a constant" transformation is the most common new-equation question type in IBPS PO.
Solving path: From x² - 7x + 10 = 0: α + β = 7, αβ = 10. New roots: α + 2 and β + 2. New sum = 7 + 4 = 11. New product = αβ + 2(α + β) + 4 = 10 + 14 + 4 = 28. New equation: x² - 11x + 28 = 0.
Why this question: The equal-roots condition (D = 0) in parametric form is a staple across bank exams.
Solving path: Equal roots require D = 0. Here a = 1, b = k, c = 9. So k² - 4(1)(9) = 0 → k² = 36 → k = ±6.
Why this question: α³ + β³ forces you to use the cubic sum identity, which collapses a hard computation into two arithmetic steps.
Solving path: α + β = 6, αβ = 8. Apply α³ + β³ = (α + β)³ - 3αβ(α + β) = 216 - 3(8)(6) = 216 - 144 = 72.
Why this question: Straight factorization — the baseline skill everything else builds on.
Solving path: Need two numbers multiplying to 6 and adding to -5: that is -2 and -3. So (x - 2)(x - 3) = 0, giving x = 2 or x = 3.
Why this question: Pure discriminant calculation — tests sign handling with a negative b.
Solving path: a = 3, b = -4, c = 1. D = (-4)² - 4(3)(1) = 16 - 12 = 4. The negative sign on b does not matter once you square it.
Why this question: Forming an equation from conjugate surd roots tests whether you use the difference-of-squares shortcut for the product.
Solving path: Sum = (2 + √3) + (2 - √3) = 4. Product = (2)² - (√3)² = 4 - 3 = 1. Equation: x² - 4x + 1 = 0.
Sign flip on sum of roots. For x² - bx + c = 0, the sum α + β = +b, not -b. The formula is α + β = -(coefficient of x) / (leading coefficient). Students in a hurry drop the negative of a negative and get the sum wrong.
Forgetting to divide by a in Vieta's formulas. When the equation is not monic (a ≠ 1), the sum is -b/a and the product is c/a. For 3x² - 7x + 2 = 0: sum = 7/3, product = 2/3 — not 7 and 2.
Middle-term split with wrong sign. When factorizing x² + 5x + 6 = 0, students sometimes write (x + 2)(x + 3) but leave the roots as +2, +3 instead of -2, -3. The roots from (x + 2)(x + 3) = 0 are x = -2 and x = -3. Always equate each factor to zero.
Using the quadratic formula when factorization would work. The formula is slow. For exam conditions, if you cannot spot the factors in 10 seconds, try listing factor pairs of ac systematically. Reach for the formula only after that.
Incorrectly expanding (α + 2)(β + 2) in root transformation. The expansion is αβ + 2α + 2β + 4 = αβ + 2(α + β) + 4, not just αβ + 4. The middle term 2(α + β) is routinely dropped.
Concluding "cannot be determined" too quickly in root comparison. Before writing CND, verify: have you correctly found all roots of both equations? A sign error in factorization can make two equal roots look unequal, pushing you toward CND when a definite answer exists.