A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a ≠ 0. The word "quadratic" comes from the Latin quadratus, meaning square — the defining feature is that the highest power of the variable is 2.
Here is the important thing to understand intuitively: a quadratic equation always has exactly two roots (counting multiplicity), and those roots can be real or complex. For competitive exams, you almost exclusively deal with real roots, so your toolkit is: factorisation, the quadratic formula, and discriminant analysis.
Think of it this way — a quadratic is just a parabola. The roots are the x-intercepts of y = ax² + bx + c. If the parabola crosses the x-axis at two points, you have two distinct real roots. If it just touches the axis, you have one repeated root. If it never touches the axis, there are no real roots. This visual map directly corresponds to the discriminant D = b² - 4ac:
D > 0 → two distinct real rootsD = 0 → one repeated (equal) rootD < 0 → no real rootsThe analogy that works best: imagine you are throwing a cricket ball. If you throw it hard enough (large positive c, small a), it stays airborne and never hits the ground in the playing range — that is the D < 0 scenario. If it just grazes the ground once, D = 0. Normal trajectories that land and bounce — that is D > 0.
For RBI Grade B, quadratics appear in two flavours: word problems (consecutive numbers, reciprocal sums, products) and symbolic manipulation (discriminant conditions, α-β expressions). Both are mechanical once you have the pattern recognition down. The discriminant questions in particular are almost free marks if you know the three conditions cold.
Given ax² + bx + c = 0:
Sum of roots: α + β = -b/a
Product of roots: αβ = c/a
Quadratic formula: x = (-b ± √(b² - 4ac)) / 2a
Don't reach for the quadratic formula as your first move in an exam. Use it only when factorisation fails. Factorisation is faster by at least 30 seconds on a clean integer-root problem.
For x² + bx + c = 0 (when a = 1): you need two numbers whose sum is b (the coefficient of x) and whose product is c (the constant). This is the Sum-Product method.
For ax² + bx + c = 0 (when a ≠ 1): use the Split-Middle method.
a × c.b and whose product is a × c.Example: 6x² - 13x + 6 = 0
a × c = 6 × 6 = 36-13 with product 36: those are -9 and -4.6x² - 9x - 4x + 6 = 03x(2x - 3) - 2(2x - 3) = 0(3x - 2)(2x - 3) = 0x = 2/3 or x = 3/2The discriminant D = b² - 4ac controls root nature. These questions are among the fastest in the quant section if you have the conditions memorised:
| Condition | Root Nature |
|---|---|
| D > 0 | Two distinct real roots |
| D = 0 | Two equal (repeated) real roots |
| D < 0 | No real roots (complex) |
For "equal roots" questions: set b² - 4ac = 0 and solve for the unknown parameter.
For "no real roots" questions: set b² - 4ac < 0 and solve the inequality.
This is where RBI Grade B separates itself from SSC-type exams. You need to evaluate expressions involving roots without actually finding the roots. The key identities:
α² + β² = (α + β)² - 2αβα³ + β³ = (α + β)³ - 3αβ(α + β)α² - β² = (α + β)(α - β), and (α - β)² = (α + β)² - 4αβα³ - β³ = (α - β)(α² + αβ + β²)For x² - 5x + 6 = 0: α + β = 5, αβ = 6.
To find α³ + β³:
= (α + β)³ - 3αβ(α + β) = 5³ - 3(6)(5) = 125 - 90 = 35
Look — you never needed to find that α = 2, β = 3 explicitly. Straight from sum-product to the answer in two lines. That is the RBI Grade B way of doing this.
Most word problems reduce to one of three frames:
Frame 1 — Consecutive integers:
x, x+1x, x+2ax² + bx + c = 0.Frame 2 — Reciprocal/sum:
x + 1/x = k → multiply by x → x² - kx + 1 = 0.Frame 3 — Area/perimeter/product:
Always check: does the problem context require a positive root? Negative values are mathematically valid roots but often physically meaningless (you cannot have a negative length or a negative count of people). Eliminate the extraneous root explicitly — examiners sometimes include it as a trap option.
When a = 1, scan the constant c and find factor pairs. Pick the pair that sums to b. This takes under 10 seconds for numbers under 100. Example: x² - 6x + 8 = 0. Factors of 8: (1,8), (2,4). Which pair sums to 6? (2,4). So roots are 2 and 4. Standard quadratic formula: ~45 seconds. Sum-Product scan: ~8 seconds. Saves roughly 37 seconds per question.
Never solve for individual roots when asked for expressions like α² + β² or α³ + β³. Read off α + β = -b/a and αβ = c/a directly, then chain the identities. For α³ + β³, you need exactly 3 multiplications and 1 subtraction. Finding roots individually and cubing them: 6+ arithmetic operations with risk of error. Vieta chain: 3 operations, zero ambiguity.
When asked "for what value of k does the equation have equal/no real roots", substitute into b² - 4ac = 0 (for equal roots) or < 0 (no real roots) immediately. Do not attempt to factor or find roots. Example: kx² + 4x + 1 = 0, equal roots → 4² - 4(k)(1) = 0 → 16 = 4k → k = 4. Total: 2 steps, under 15 seconds. Trying to factor with unknown k: impossible without this condition first.
For "sum of squares of two consecutive odd numbers = 130" type questions, estimate first. √(130/2) ≈ 8. So the numbers are near 8. Odd numbers near 8: try 7 and 9. 49 + 81 = 130. Done. This estimation-then-verify approach takes under 20 seconds. Setting up and solving the full quadratic: ~60 seconds. Reserve the algebraic method only when estimation gives non-integers.
"Sum of a number and its reciprocal = k" always gives the symmetric form x² - kx + 1 = 0. The product of roots is always 1 (one root is the reciprocal of the other). So if one root is p/q, the other is automatically q/p. You only need to find one root; the other is its reciprocal. Halves the checking work.
When you encounter a quadratic question in the exam hall, run this decision tree:
Step 1 — Classify the question type.
Step 2 — Choose the solving method.
a = 1 and c is a small integer: try Sum-Product first (under 10 seconds).a ≠ 1: use Split-Middle (multiply a × c, find factor pair summing to b).Step 3 — Sanity check.
α + β = -b/a and αβ = c/a? Two-second cross-check.Step 4 — For inequalities on k: after setting D < 0 or D > 0, double-check the direction of the inequality sign — sign flips when dividing by a negative coefficient.
Why this question: This is the canonical reciprocal-sum problem. It tests whether you can translate the word problem and then factor a non-monic quadratic.
Solving path: Let the number be x. The condition "sum of a number and its reciprocal is 13/6" translates to x + 1/x = 13/6. Multiply both sides by 6x: 6x² + 6 = 13x, giving 6x² - 13x + 6 = 0. Now apply Split-Middle: a × c = 36, need factor pair summing to -13: that is -9 and -4. Split: 6x² - 9x - 4x + 6 = 0 → 3x(2x - 3) - 2(2x - 3) = 0 → (3x - 2)(2x - 3) = 0. Roots: x = 2/3 or x = 3/2. Note that one root is the reciprocal of the other — exactly as the Symmetric Quadratic trick predicts.
Why this question: The simplest factorisation case — every exam has at least one of these. Demonstrates the Sum-Product method in its cleanest form.
Solving path: x² - 6x + 8 = 0. Factors of 8 that sum to 6: (2, 4). So (x - 2)(x - 4) = 0, giving roots 2 and 4. Cross-check: 2 + 4 = 6 = -(-6)/1 and 2 × 4 = 8 = 8/1. Both verified in one glance.
Why this question: Tests the consecutive-odd-numbers frame and whether you can handle an equation that reduces to a clean factorisable form. The estimation trick works beautifully here.
Solving path: Estimate first — √(130/2) ≈ 8.06. Odd numbers near 8: try 7 and 9. 49 + 81 = 130. Confirmed. If you prefer algebra: let the numbers be x and x + 2. Then x² + (x+2)² = 130 expands to 2x² + 4x + 4 = 130, so x² + 2x - 63 = 0. Factor: (x + 9)(x - 7) = 0. Roots: x = 7 or x = -9. Since we need positive odd numbers, x = 7.
Why this question: The equal-roots discriminant condition is a high-frequency RBI pattern. Tests whether you immediately reach for D = 0 rather than trying to factor with an unknown parameter.
Solving path: Equal roots condition: D = b² - 4ac = 0. Here a = k, b = 4, c = 1. So 16 - 4k = 0, giving k = 4. Direct. No factorisation attempted, no quadratic formula used. Two arithmetic steps.
Why this question: This is the most RBI Grade B-flavoured question in the set — it tests Vieta's relations and the cubic sum identity. You never need to find that the roots are 2 and 3.
Solving path: From x² - 5x + 6 = 0: α + β = 5 (Vieta: -b/a = 5/1), αβ = 6 (Vieta: c/a = 6/1). Apply the identity α³ + β³ = (α + β)³ - 3αβ(α + β) = 5³ - 3(6)(5) = 125 - 90 = 35. Done in three arithmetic operations. Option (a) is 35.
Forgetting to reject extraneous roots in word problems. When you get x = 12 or x = -14 for "consecutive even numbers", the answer is 12. But watch the question — if it asks for "the numbers" (plural) and both are positive by context, you need both. Read the question stem twice before writing the final answer.
Sign error in Vieta's sum formula. α + β = -b/a, not +b/a. This is the single most common error in α-β questions. For x² - 5x + 6 = 0, the sum is +5 because -(-5)/1 = 5. Students who memorise it as "coefficient of x" without the negative sign will get α³ + β³ = (-5)³ - ... = -125 - ... — a wildly wrong answer.
Using the quadratic formula when factorisation is faster. The formula is a fallback, not the primary method. On a clean problem like x² - 6x + 8 = 0, the formula takes 40+ seconds and introduces arithmetic errors. Sum-Product takes 8 seconds. Develop the reflex to try factorisation first.
Discriminant inequality direction. For "no real roots" of 3x² - 2x + k = 0, you get 4 - 12k < 0, so k > 1/3. Many students write k < 1/3 because they divide both sides by -12 and forget to flip the inequality sign. Specifically: when you divide or multiply an inequality by a negative number, the sign flips.
Applying Sum-Product to the wrong coefficient. For ax² + bx + c = 0 with a ≠ 1, the sum of roots is -b/a and the product is c/a — not simply -b and c. When a = 1, the formulas simplify because dividing by 1 changes nothing, and students who learned the pattern with a = 1 examples then misapply it to 6x² - 13x + 6 = 0.
Consecutive even vs consecutive odd — using x, x+1 instead of x, x+2. Consecutive even numbers are x and x+2 (e.g., 12 and 14). Consecutive odd numbers are also x and x+2 (e.g., 7 and 9). Using x and x+1 gives you consecutive integers, not consecutive even or odd integers. This produces a completely different equation and a wrong answer.