A quadratic equation is any equation of the form ax² + bx + c = 0, where a ≠ 0. The word "quadratic" comes from the Latin for "square" — the defining feature is that the highest power of the variable is 2.
Here is a grounded way to think about it: a linear equation gives you one unknown and one root. A quadratic gives you one unknown but two roots — because squaring a number destroys the sign information, so two different inputs (+3 and -3) can produce the same output (9). Every quadratic equation therefore has exactly two roots, though they may be equal to each other, or non-real (complex).
In IBPS RRB PO, quadratic equations appear in two distinct formats:
Format 1 — Direct solve. You get an equation like x² - 7x + 12 = 0 and must find the roots.
Format 2 — Relationship questions. You get something like "if α and β are roots of x² - 3x + 1 = 0, find α² + β²". This tests whether you can derive quantities from roots without actually computing the roots themselves.
Both formats rely on the same two weapons: Vieta's formulas (the sum-product relationships) and factorisation. Master those two, and the discriminant is just a supporting check.
Think of it this way: the equation ax² + bx + c = 0 is a container. The coefficients a, b, c are not random — they encode the roots. Vieta's formulas are the key that decodes them. The moment you see a quadratic, your first instinct should be: "What does b tell me about the sum, and what does c tell me about the product?"
For ax² + bx + c = 0 with roots α and β:
These are non-negotiable. Write them on your palm if you must. The logic behind them: if α and β are roots, then:
Matching coefficients gives you Vieta's formulas directly.
Important sign trap: The sum formula is -b/a, not b/a. For x² - 7x + 12 = 0, b = -7, so sum = -(-7)/1 = 7. This sign error catches candidates every single time.
For most IBPS RRB PO questions, the quadratic will factor cleanly. The approach:
Given x² + bx + c = 0 (with a = 1), find two numbers p and q such that:
p + q = b (the coefficient of x)p × q = c (the constant term)Then the equation factors as (x - p)(x - q) = 0, giving roots p and q.
Example: x² - 4x + 3 = 0
You need two numbers that add to -4 and multiply to 3. Those are -1 and -3.
So: (x - 1)(x - 3) = 0, roots are 1 and 3.
When a ≠ 1, scale first. For 2x² - 8x + 6 = 0, factor out 2 to get 2(x² - 4x + 3) = 0, then solve x² - 4x + 3 = 0 as above.
D = b² - 4ac
| Discriminant | Nature of roots |
|---|---|
| D > 0 | Two distinct real roots |
| D = 0 | Two equal real roots (repeated root) |
| D < 0 | No real roots (complex roots) |
In exam questions, the discriminant appears most often in "find k such that roots are equal" format. Set D = 0 and solve for k. That's it — don't overthink it.
These are the "trap questions" because they look hard but are mechanical once you see the pattern. The two master identities:
From these, everything else follows. You almost never need to compute the actual values of α and β.
Example: x² - 3x + 1 = 0, find α² + β².
α + β = 3, αβ = 1.
α² + β² = (3)² - 2(1) = 9 - 2 = 7.
Done in under 10 seconds — no quadratic formula needed.
If you know the roots, you can reverse-engineer the equation:
This is the "roots → equation" direction. In the exam, you might be given roots 2 and -3 and asked to identify the equation. Sum = -1, product = -6, so the equation is x² - (-1)x + (-6) = x² + x - 6 = 0.
Some IBPS RRB PO questions give you two equations (one in x, one in y) and ask you to compare roots. The process:
x and roots of y.x > y, x < y, x ≥ y, x ≤ y, or "cannot be determined".The "cannot be determined" option applies when the ranges overlap — for instance, if x ∈ \{1, 4\} and y ∈ \{2, 3\}, some values of x exceed y and some don't.
The sum of roots is -b/a — the sign of b flips. The product is c/a — no flip. A fast mnemonic: "Sum changes sign, Product is fine." For x² - 5x + 6 = 0: sum = +5 (sign flipped from -5), product = +6 (unchanged). Standard method: re-derive from formula (15s). This pattern: read coefficients directly (3s). The sign flip on sum is the #1 source of errors in this topic — catching it saves you half a mark per question.
Before attempting any factorisation, compute target sum (-b/a) and target product (c/a) in your head. Then mentally scan factor pairs of the product and check which pair sums correctly. For x² - 8x + 15 = 0: product = 15, factor pairs are (1,15), (3,5). Sum target = 8. Pair (3,5) sums to 8 — roots are 3 and 5. Standard factorisation by trial: 30s. This scan: under 10s. It works because you narrow the search space using both conditions simultaneously rather than testing one at a time.
For any question involving α² + β², α³ + β³, or (α - β)², the entry point is always the same: compute (α + β)² first, then subtract. α² + β² = (α + β)² - 2αβ. (α - β)² = (α + β)² - 4αβ. α³ + β³ = (α + β)³ - 3αβ(α + β). You never need the actual roots. Full quadratic formula route: 60s. Identity substitution: 10s. Learn the three identities above and you handle every alpha-beta question at this level.
When a question asks "for what value of k do roots become equal/real", set b² - 4ac = 0 directly. For x² + 2x + k = 0: 4 - 4k = 0, so k = 1. Eliminate all answer choices where k would make D > 0 or D < 0. Standard route: compute discriminant for each option (40s). Direct set-to-zero: 8s. This works because "equal roots" is a precise discriminant condition, not a fuzzy one — there is exactly one value of k.
When asked to identify the quadratic with given roots, just write x² - (sum)x + (product) = 0. For roots 2 and -3: sum = -1, product = -6, equation = x² + x - 6 = 0. You don't need to expand (x-2)(x+3) at all. Expansion method: 20s with sign errors possible. Vieta reverse: 5s. Specifically useful for eliminating wrong options in multiple-choice — check only the x-coefficient sign (it equals -(sum)) and the constant term (it equals product).
When you see a quadratic question in the exam, run this decision tree:
Step 1 — Identify the question type.
α² + β² or similar" → use identity, done.Step 2 — Factor or formula?
c/a is a small integer (under 20), try factorisation first. Scan factor pairs of the product, check which pair matches the sum. 90% of IBPS-level quadratics factor cleanly.x = (-b ± √(b²-4ac)) / 2a.Step 3 — Sign check before writing the answer.
Always verify: does root 1 × root 2 equal c/a? This 2-second check catches arithmetic errors without re-solving.
Step 4 — Root comparison (if two-equation format). Solve each equation, list all roots, compare. When ranges overlap, mark "cannot be determined" — don't guess.
Why this question: Tests the most basic Vieta's formula — sum of roots. This is free marks if you remember the sign rule.
Solving path: Identify a = 1, b = -7. Sum of roots = -b/a = -(-7)/1 = 7. The equation also factors as (x-3)(x-4) = 0, confirming roots 3 and 4, sum = 7. Either route confirms option A.
Why this question: Tests factorisation with a ≠ 1. The key move is factoring out 2 first to simplify.
Solving path: Factor out 2: 2(x² - 4x + 3) = 0. Now solve x² - 4x + 3 = 0. Need two numbers that multiply to 3 and add to -4: those are -1 and -3. So (x-1)(x-3) = 0, giving x = 1 or x = 3. Answer: option A.
Why this question: Tests Vieta's product formula. One-line solution if you know c/a.
Solving path: Product of roots = c/a = k/1 = k. Given product = 6, so k = 6. Answer: option A. Took under 5 seconds — don't reach for the quadratic formula here.
Why this question: The classic alpha-beta identity question. This format appears regularly and is designed to look harder than it is.
Solving path: From x² - 3x + 1 = 0: α + β = 3, αβ = 1. Apply the identity: α² + β² = (α + β)² - 2αβ = 9 - 2(1) = 7. Answer: option A. Note — the actual roots involve irrational numbers. The identity bypasses that entirely.
Why this question: A two-step Vieta problem that looks like it needs more information than it does. Tests whether you connect sum-of-squares to the identity.
Solving path: Sum of roots = p = 4 (given directly). α² + β² = p² - 2q = 10. Substitute: 16 - 2q = 10, so 2q = 6, q = 3. Answer: option A.
Why this question: Discriminant condition for equal roots. Tests whether you can set D = 0 and isolate k.
Solving path: For equal roots, D = b² - 4ac = 0. Here a = 1, b = 2, c = k. So 4 - 4k = 0, giving k = 1. Answer: option A. Verify: x² + 2x + 1 = (x+1)² = 0 — yes, equal root at x = -1.
Why this question: Reverse Vieta — constructing an equation from its roots. Tests the formula x² - (sum)x + (product) = 0.
Solving path: Roots are 2 and -3. Sum = 2 + (-3) = -1. Product = 2 × (-3) = -6. Equation: x² - (-1)x + (-6) = x² + x - 6 = 0. Answer: option A. The sign of the middle term is the trap — -sum becomes +1x here.
Forgetting the negative sign in the sum formula. The sum of roots is -b/a, not b/a. For x² - 7x + 12 = 0, b = -7, so sum = +7 — not -7. This single error accounts for a large fraction of wrong answers on Vieta questions. Say it out loud: "b flips, c stays."
Not factoring out the leading coefficient first. For 2x² - 8x + 6 = 0, treating it as if a = 1 and jumping to factor pairs of 6 that sum to -8 will fail. Always divide through by a (or factor it out) before scanning for factor pairs.
Using the quadratic formula when factorisation is faster. The formula is a fallback, not a first resort. On clean integer-coefficient questions (which is almost all of IBPS), factorisation takes under 15 seconds. The formula takes 40+ seconds and introduces calculation errors under pressure.
Mixing up α² + β² with (α + β)². These are not the same. (α + β)² = α² + 2αβ + β². You must subtract 2αβ to get α² + β². Drawing this expansion once on scratch paper cements it permanently.
Root comparison "cannot be determined" avoidance. Many candidates refuse to mark "cannot be determined" because it feels like giving up. But when the root sets overlap (e.g., x ∈ \{1, 5\} and y ∈ \{2, 4\}), it is the correct answer. If some roots satisfy x > y and others satisfy x < y, the relationship is genuinely indeterminate.
Sign errors when constructing equations from roots. The equation is x² - (α + β)x + αβ = 0. The sum appears with a negative sign in front of it. For roots 2 and -3 where sum = -1, the equation is x² - (-1)x + (-6) = x² + x - 6. Writing x² - x - 6 (positive sum without the double-negative correction) is one of the most common wrong answers in this format.