A quadratic equation is any equation of the form ax² + bx + c = 0 where a ≠ 0. That last condition matters — the moment a = 0, you've got a linear equation, and all the theory below collapses.
Here's a clean way to think about it: a quadratic is just a parabola intersecting the x-axis. The roots are the x-coordinates of those intersection points. When the parabola cuts the x-axis at two points, you get two distinct real roots. When it just touches, you get one repeated root. When it floats entirely above or below the x-axis without touching, the roots are complex (imaginary).
That visual — parabola and x-axis — is your anchor for everything that follows.
Why quadratics matter for NDA specifically: NDA Paper-I (Mathematics) tests quadratics not just as standalone questions but as the backbone of several other topics. Nature of roots connects to complex numbers. Sum-product formulas appear in progressions questions (roots in AP/GP). Completing the square is the gateway to conic sections. You are going to see this topic folded into at least 3-4 other chapters, so the investment here compounds.
The standard form ax² + bx + c = 0 gives roots:
The expression b² - 4ac is called the discriminant (D). It controls everything about the nature of roots:
D > 0: two distinct real rootsD = 0: two equal real roots (repeated root)D < 0: two complex conjugate rootsThink of D as the "mood indicator" of the equation. Before you even attempt to find the roots, read D — it tells you what kind of answer to expect.
For ax² + bx + c = 0 with roots α and β:
These are not just formulas to memorize — they're the primary tool for NDA questions. Most quadratic problems that look computational are actually Vieta problems in disguise. Look for phrases like "roots exceeding each other by," "arithmetic mean of roots," "product of roots equals sum" — these all translate immediately into Vieta expressions.
Forming a quadratic from known roots: If you know α + β and αβ, the equation is:
This is the reverse application of Vieta. NDA loves asking you to construct equations, especially when roots are given in terms of another equation's roots.
For the equation ax² + bx + c = 0 where a, b, c ∈ ℝ:
| D value | Nature of roots |
|---|---|
| D > 0 and a perfect square | Two distinct rational roots |
| D > 0 and not a perfect square | Two distinct irrational roots (conjugate surds) |
| D = 0 | Equal real roots: α = β = -b/2a |
| D < 0 | Complex conjugate roots: p ± qi |
When a, b, c ∈ ℚ (rational coefficients), irrational and complex roots always come in conjugate pairs. So if 2 + √3 is a root, 2 - √3 is automatically the other root. If 3 + 2i is a root, 3 - 2i is the other.
Completing the square converts ax² + bx + c into vertex form a(x + h)² + k. This is essential for two things: finding the minimum/maximum value of a quadratic expression, and solving equations when factorization isn't obvious.
Method: Take ax² + bx + c, divide through by a, then:
The minimum value of the expression (for a > 0) is (4ac - b²) / 4a, achieved at x = -b/2a.
Note the sign carefully: minimum = (4ac - b²)/4a = -D/4a. This is a direct formula NDA has tested.
When a question asks "for what values of k are roots real?" — set D ≥ 0 and solve the resulting inequality in k. When it asks "for what values of k are roots equal?" — set D = 0. This is the standard parametric approach.
A particularly elegant result: if a, b, c are in GP, then b² = ac. Substituting into the discriminant: D = b² - 4ac = ac - 4ac = -3ac. Since a, c > 0 (GP with positive terms), D = -3ac < 0 — roots are always imaginary. This appeared directly in an NDA PYQ.
Equations like √x + x + 1 = 0 or 5√x - x - 4 = 0 look non-standard but become quadratics with the substitution y = √x (so y ≥ 0). After solving for y, remember to check: y = √x ≥ 0, so discard negative y values. Then x = y².
This is a common NDA trap — the substitution is straightforward, but students forget the constraint y ≥ 0 and report roots that don't exist in ℝ⁺.
If roots \{α, β\} must equal \{α², β²\}, there are two structural cases:
α = α² means α ∈ \{0, 1\}.α = β² and β = α², which gives α = α⁴, so α³ = 1 with α ≠ 1, making α a primitive cube root of unity.Counting carefully: fixed-point pairs give (0,0), (0,1), (1,1) — three equations. The swap case gives one equation (x² + x + 1 = 0). Total: 4.
Before computing roots, always compute D = b² - 4ac. If D < 0, you know roots are complex — write the answer immediately without grinding through the formula. If D is a perfect square, roots are rational and factorization will be faster than the formula. This single habit saves 30-45 seconds per question by routing you to the faster method before you start computing.
Micro-example: x² + x + 1 = 0. D = 1 - 4 = -3 < 0. Stop here — roots are complex. Standard method (formula application): ~40 seconds. Discriminant check: 8 seconds.
When a question gives you two equations sharing roots, use Vieta instead of solving. If α, β are roots of equation 1, read off α+β and αβ in one step. Then answer questions about a second equation by substituting into its Vieta relationships.
Micro-example: Roots of x² - 2bx + c² = 0 have α+β = 2b and αβ = c². So A = b, G = c. The equation x² - (b+c)x + bc = 0 has sum b+c = A+G and product bc = A·G — it factors to (x-b)(x-c) = 0, roots A and G. This takes 4 steps vs. 10+ steps if you solve explicitly for α and β.
For ax² + bx + c with a > 0, minimum value = (4ac - b²) / 4a = -D/4a. No need to complete the square from scratch every time.
Micro-example: x² + kx + k². Here a=1, b=k, c=k². Minimum = (4·1·k² - k²) / 4 = 3k²/4. Standard completing-the-square method: 5 steps, ~35 seconds. Direct formula: 2 steps, ~10 seconds.
(α - β)² = (α + β)² - 4αβ = (b/a)² - 4(c/a) = D/a²
So |α - β| = √D / a.
When a question says "one root exceeds the other by k," set √D / a = k and solve. Don't parametrize with individual roots.
Micro-example: x² - kx + k = 0, one root exceeds other by 2√3. So (2√3)² = D/1, giving 12 = k² - 4k, then k² - 4k - 12 = 0, (k-6)(k+2) = 0. This takes 3 steps from the formula. Setting up α and β individually takes 6+ steps.
If a, b, c are in GP (positive terms), the roots of ax² + bx + c = 0 are always imaginary. Proof shortcut: D = b² - 4ac = ac - 4ac = -3ac < 0. Commit this result — NDA has tested it and will again. Recognition time: 3 seconds vs. 25 seconds to re-derive.
When you see a quadratic question in the exam hall, run this decision tree:
Step 1 — Identify what's being asked. Is it (a) find the roots, (b) nature of roots, (c) a condition on roots, or (d) minimum/maximum of expression?
Step 2 — Compute D immediately for (a) and (b). D routes you: rational roots → factorize; irrational → formula; complex → write p ± qi form directly.
Step 3 — For (c) conditions on roots: translate immediately to Vieta. "Roots in ratio 1:2" → let roots be r, 2r, then 3r = -b/a and 2r² = c/a. Never solve the equation; always work with sum and product.
Step 4 — For (d) minimum/maximum: use (4ac - b²)/4a directly for the vertex value. For min of a sum-type expression, check if completing the square is faster.
Step 5 — Substitution check: if the equation has √x or x^(1/2), substitute y = √x, enforce y ≥ 0, solve, then square back.
One meta-rule: the quadratic formula is a last resort for NDA, not a first move. Factorization and Vieta together solve 80% of quadratic questions faster.
Why this question: Tests whether you can enumerate the structural cases systematically rather than guessing.
Solving path: Set up the condition: squaring roots must return the same set \{α, β\}. This has exactly two structural sub-cases. Case 1 (fixed points): α = α² means α(α-1)=0, so α ∈ \{0,1\}. Pairs from \{0,1\}: (0,0), (0,1), (1,1) — giving three monic quadratics. Case 2 (swap): α = β² and β = α². Substituting: α = α⁴, so α³ = 1, α ≠ 1 (since Case 2 requires the roots to genuinely swap). This gives the primitive cube roots of unity, yielding the single equation x² + x + 1 = 0. Total: 3 + 1 = 4.
Why this question: Connects two equations through AM and GM — tests whether you recognize Vieta relationships across equations without solving either.
Solving path: From x² - 2bx + c² = 0: sum of roots = 2b, product = c². So A = (α+β)/2 = b and G = √(αβ) = √(c²) = c (since b, c > 0). Now the second equation x² - (b+c)x + bc = 0: its sum is b + c = A + G and product is bc = A·G. Factor: (x-b)(x-c) = 0, roots are b and c, which are exactly A and G.
Why this question: Minimum value of a quadratic expression — a direct test of the completing-the-square or vertex formula.
Solving path: Apply the minimum formula directly. For x² + kx + k², we have a=1, b=k, c=k². Minimum = (4·1·k² - k²)/(4·1) = 3k²/4. Alternatively, complete the square: (x + k/2)² + k² - k²/4 = (x + k/2)² + 3k²/4. Minimum is 3k²/4 at x = -k/2.
Why this question: Disguised quadratic via substitution — tests whether you handle the domain constraint y = √x ≥ 0.
Solving path: Statement I: √x + x + 1 = 0. Let y = √x ≥ 0. Then y + y² + 1 = 0, i.e., y² + y + 1 = 0. Discriminant = 1 - 4 = -3 < 0. No real y, so no real x. Statement I is false (there are no real roots at all, let alone irrational ones). Statement II: 5√x - x - 4 = 0. Let y = √x ≥ 0. Then -y² + 5y - 4 = 0, i.e., y² - 5y + 4 = 0, (y-1)(y-4) = 0, y = 1 or y = 4. Both y ≥ 0, so x = 1 or x = 16. Both rational. Statement II is correct.
Why this question: Classic "difference of roots" setup — tests the shortcut (α-β)² = D/a².
Solving path: For x² - kx + k = 0: sum = k, product = k. Use the identity (α-β)² = (α+β)² - 4αβ = k² - 4k. Given |α-β| = 2√3, so (α-β)² = 12. Therefore k² - 4k = 12, giving k² - 4k - 12 = 0, (k-6)(k+2) = 0, k = 6 or k = -2. Among the options, k = 6.
Why this question: Tests the GP-in-coefficients result — discriminant behavior plus root ratios involving cube roots of unity.
Solving path: Since a, b, c are in GP with a, c > 0, we have b² = ac. Statement 1: D = b² - 4ac = ac - 4ac = -3ac < 0. Roots are imaginary. Correct. Statement 3: Product of roots = c/a. Since b² = ac, c = b²/a, so c/a = b²/a². Correct. Statement 2: The roots are (-b ± i√(3ac))/(2a). Since √(ac) = |b|, roots become (-b ± i√3·|b|)/(2a). Their ratio is (-b + i√3|b|)/(-b - i√3|b|). Factoring out |b| (assuming b > 0, so |b| = b): ratio = (-1 + i√3)/(-1 - i√3) = e^{i·2π/3}/e^{-i·2π/3} which equals ω/ω² = ω^{-1}, or equivalently the ratio is 1:ω. Correct. All three correct.
Forgetting a ≠ 0. When a parameter multiplies x², always check whether the parameter can be zero. If it can, the equation degenerates to linear and the quadratic theory doesn't apply.
Dropping the domain on substituted variables. In y = √x substitutions, y ≥ 0 is mandatory. Negative y solutions correspond to no real x and must be discarded. This mistake directly flips the answer on questions like Statement I above.
Confusing D < 0 with "no solution." D < 0 means no real roots — there are still two complex roots. If a question asks about complex roots or cube roots of unity, D < 0 is actually the starting point, not a dead end.
Using the quadratic formula when Vieta suffices. NDA questions framed as "if α, β are roots, find α² + β²" don't need you to find α and β explicitly. Use α² + β² = (α+β)² - 2αβ = (b/a)² - 2(c/a). Solving the formula first wastes 40-60 seconds.
Sign errors in Vieta's formulas. The sum of roots is -b/a, not +b/a. Under time pressure, this sign flip is the most common arithmetic error. Write "sum = -b/a" explicitly at the top of your working, not from memory mid-calculation.
Treating "equal roots" and "real roots" as the same condition. Equal roots require D = 0; real roots require D ≥ 0. When a question asks for the condition that roots are real and equal, both conditions collapse to D = 0 — but when it just asks "real," setting D = 0 instead of D ≥ 0 will give you the wrong answer.