A quadratic equation is any equation of the form ax² + bx + c = 0 where a ≠ 0. The highest power of the variable is 2 — that is the defining feature. When a = 0, it collapses to a linear equation, so the condition a ≠ 0 is not optional.
Here is the analogy that makes this click: think of a quadratic as a parabola — a U-shaped (or inverted-U) curve. The "roots" are simply the x-coordinates where the parabola cuts the x-axis. If the parabola cuts in two places, you have two distinct roots. If it just touches the x-axis at one point, you have two equal roots. If it never reaches the x-axis at all, you have no real roots. Every nature-of-roots question you will ever see in SBI PO is just asking you to figure out which of these three situations you are in — and the discriminant D = b² - 4ac is the single tool that answers it.
For the exam, quadratic equations show up in three flavors. First, straightforward solving — find the roots. Second, Vieta's relations — given a property of the roots, find an unknown coefficient. Third, quadratic inequalities — given f(x) > 0 or f(x) < 0, find the range of x. The third type is where most people lose time because they forget the sign-analysis step. This page addresses all three with a repeatable process.
One subtle variant that appears in SBI PO but is often unprepared for: equations involving |x| (absolute value). The trick there is a substitution — let y = |x|, solve the resulting quadratic in y, then unpack the absolute value at the end. You will see a live example in the PYQ section.
Given ax² + bx + c = 0, the two roots are:
x = (-b ± √(b² - 4ac)) / (2a)
This is the quadratic formula. In the exam hall, reach for this formula only as a last resort — factorization is almost always faster when the numbers are clean.
Look — this is the highest-leverage concept in quadratic equations for banking exams. If α and β are the two roots of ax² + bx + c = 0, then:
α + β = -b/aαβ = c/aFor the specific case x² + bx + c = 0 (leading coefficient = 1, which is the most common exam form):
α + β = -bαβ = cAnd critically, the reverse: if you know the roots, you can construct the equation as:
x² - (sum)x + (product) = 0
This is the construction formula — memorize it in that exact form. The coefficient of x has a minus sign; the constant term has a plus sign. This is where people make sign errors under pressure.
D = b² - 4ac
| Condition | Nature of roots |
|---|---|
| D > 0 | Two distinct real roots |
| D = 0 | Two equal real roots (repeated root) |
| D < 0 | No real roots (two complex conjugate roots) |
For SBI PO, you will also see conditions phrased as inequalities — "for what values of k does the equation have real and distinct roots?" That is just asking you to set D > 0 and solve for k.
For ax² + bx + c = 0, the split-middle-term method works as follows:
a × c (product of leading and constant coefficient).ac and whose sum is b.Example — 2x² - 5x + 3 = 0:
a × c = 2 × 3 = 62x² - 3x - 2x + 3 = 0x(2x - 3) - 1(2x - 3) = 0 → (2x - 3)(x - 1) = 0x = 3/2 or x = 1This is the method. Not the quadratic formula. The formula is a 40-second process; clean factorization is a 15-second process when you have integer or simple fractional roots — which is almost always in SBI PO.
These come up in "find α² + β²" type questions. Do not re-derive from scratch each time:
α² + β² = (α + β)² - 2αβ(α - β)² = (α + β)² - 4αβ — this gives you the difference of rootsα³ + β³ = (α + β)³ - 3αβ(α + β)The first identity is the most tested. Commit the pattern to memory: square the sum, subtract twice the product.
For f(x) = (x - α)(x - β) > 0 where α < β:
x² is positive).f(x) > 0 outside the roots: x < α or x > β, i.e., x ∈ (-∞, α) ∪ (β, ∞)f(x) < 0 between the roots: α < x < β, i.e., x ∈ (α, β)The rule to remember: positive outside, negative inside (when the leading coefficient is positive). If the leading coefficient is negative, flip the signs — the parabola opens downward.
When the inequality is ≥ or ≤, include the endpoints (closed brackets). When it is strict > or <, exclude them (open brackets).
When you see x² alongside |x|, substitute y = |x| where y ≥ 0. Solve the resulting equation in y, then recover x. Since |x| = y means x = ±y, you get two x-values for each positive solution in y. If y = 0, you get only x = 0.
When you are given roots and asked for the equation, the template is always:
x² - (S)x + (P) = 0
where S = sum of roots and P = product of roots. The minus sign before S is fixed — it never changes regardless of whether the sum is positive or negative. The plus sign before P holds even if the product is negative (a negative P just makes it subtraction in practice).
Micro-example: roots are 3 and 4. S = 7, P = 12. Equation: x² - 7x + 12 = 0. Verify: (x-3)(x-4) = x² - 7x + 12. Done in 8 seconds versus expanding and checking coefficients separately (25 seconds).
For any question asking α² + β² given a quadratic, never compute the roots individually. Use:
α² + β² = (α + β)² - 2αβ
Read off α + β = -b/a and αβ = c/a directly from the equation. The calculation becomes purely arithmetic.
Micro-example: x² - 6x + 8 = 0. Sum = 6, product = 8.
α² + β² = 6² - 2(8) = 36 - 16 = 20. Total time: 10 seconds.
Finding roots first (2 and 4) then squaring and adding: 2² + 4² = 4 + 16 = 20 — also works, but takes 25 seconds. The identity scales better when roots are irrational.
Any question of the form "find k given equal roots" is a one-liner: set D = 0.
D = b² - 4ac = 0 → b² = 4ac
Micro-example: x² - px + 36 = 0, equal roots. Here b = -p, a = 1, c = 36.
p² = 4(1)(36) = 144 → p = ±12. Three arithmetic steps. Attempting to factor or use the formula without this insight takes 60+ seconds and risks sign errors.
For a monic quadratic (leading coefficient +1) with roots α and β (α < β):
So (x-2)(x-3) > 0 is satisfied outside: x < 2 or x > 3.
And (x-2)(x-3) < 0 is satisfied inside: 2 < x < 3.
Write "OUT is positive, IN is negative" at the top of your rough sheet. Eliminates the need to test values every time. Standard test-value approach: 45 seconds. This mental rule: 8 seconds once the roots are found.
Any time you see both x² and |x| in the same equation, substitute y = |x|. Because x² = |x|²= y², the equation converts to a clean quadratic in y.
Solve for y ≥ 0, then recover x = ±y for each positive solution.
Micro-example: x² - 4|x| + 3 = 0 → y² - 4y + 3 = 0 → (y-1)(y-3) = 0 → y = 1 or y = 3 → x = ±1 or x = ±3.
Without this substitution, attempting to split into cases x ≥ 0 and x < 0 separately takes 70+ seconds and is error-prone. This substitution reduces it to a standard 15-second factorization.
In the exam hall, look at the question type first before doing any calculation:
Step 1 — Classify the question.
α² + β² or similar expression" → go to the derived identity, not the roots.f(x) > 0" → find roots, then apply the parabola sign rule.|x| → substitute y = |x| before anything else.Step 2 — Extract S and P from the equation before touching the options.
For ax² + bx + c = 0: S = -b/a, P = c/a. Write these down. Half the question is already answered.
Step 3 — Verify sign on the construction formula.
If building an equation from roots: x² - (S)x + (P) = 0. The minus before S is a trap. The options will always include a version with +S to catch this error.
Step 4 — Check the discriminant condition.
"Real roots" → D ≥ 0. "Distinct real roots" → D > 0. "Equal roots" → D = 0. Never mix these up under time pressure.
Why this question: This tests the absolute-value substitution — a variant that trips up anyone who hasn't seen it in the right framing.
Solving path: Recognize the |x| term. Substitute y = |x| → y² - 4y + 3 = 0 → (y-1)(y-3) = 0 → y = 1 or y = 3. Since y = |x|, this gives x = ±1 or x = ±3. Answer: ±1, ±3.
Why this question: A pure Vieta's identity question. The moment you see α² + β², reach for the identity — not the roots.
Solving path: From x² - 6x + 8 = 0: S = 6, P = 8. Apply α² + β² = S² - 2P = 36 - 16 = 20. Done. No need to find that the roots are 2 and 4.
Why this question: Tests the construction formula in reverse — given roots, recover coefficients. The sign trap is embedded in the options.
Solving path: Equation form is x² - px + q = 0. S = 2 + 3 = 5 = p. P = 2 × 3 = 6 = q. The equation x² - px + q = 0 means -p = -(sum) → p = sum = 5, and q = product = 6.
Why this question: Straightforward factorization but with a non-unity leading coefficient. Tests whether you use the split-middle-term method correctly.
Solving path: 2x² - 5x + 3 = 0. Product = 2×3 = 6, sum = -5. Split: -3 and -2. Rewrite: 2x² - 3x - 2x + 3 = x(2x-3) - 1(2x-3) = (2x-3)(x-1) = 0. Roots: x = 3/2 or x = 1.
Why this question: Tests the quadratic inequality workflow — factorize, identify roots, apply the parabola sign rule.
Solving path: x² - 5x + 6 = (x-2)(x-3). Roots are 2 and 3. For the product to be positive (> 0), both factors must have the same sign. Apply the parabola sign rule: the expression is positive outside the roots. Therefore x < 2 or x > 3, which is (-∞, 2) ∪ (3, ∞).
Wrong sign on the sum in the construction formula. The equation is x² - (S)x + (P) = 0, not x² + (S)x + (P) = 0. If S is positive, the coefficient of x is negative. This is the single most tested trap in multiple-choice options — the wrong-sign version is always present.
Confusing D > 0 with D ≥ 0. "Real and distinct" requires D > 0 (strictly). "Real roots" (which includes equal roots) requires D ≥ 0. Writing D ≥ 0 when the question says "distinct" will give you the wrong inequality boundary.
Including the root points in a strict inequality. When the question says f(x) > 0, the answer uses open brackets at the roots. Writing (-∞, 2] ∪ [3, ∞) instead of (-∞, 2) ∪ (3, ∞) is wrong even though the logic is right.
Forgetting the ± when unpacking |x| = y. After substituting y = |x| and finding y = 3, you get x = +3 AND x = -3. Missing the negative root means you will never match the correct answer option, which always includes both signs.
Using the quadratic formula for clean-integer problems. This wastes 25-30 extra seconds and introduces arithmetic errors in the √(b² - 4ac) step. If the product a × c gives you a number with obvious factor pairs summing to b, always factorize.
Applying Vieta's formula with the wrong sign on the sum. For ax² + bx + c = 0, sum = -b/a, not +b/a. In x² - 6x + 8 = 0, b = -6, so sum = -(-6)/1 = 6. Writing sum = -6 and then computing (-6)² - 2(8) = 36 - 16 = 20 accidentally gives the right answer here, but the same sign error will fail you on construction-formula problems.