📘 Algebra · Chapter MN03🎯 NDA Level : High Priority
Quadratic equations are among the most consistently tested topics in NDA Mathematics. Questions test the discriminant, sum and product of roots, formation of new equations, and solving inequalities — all with minimal calculation and maximum conceptual clarity. Mastering this chapter gives you reliable, fast marks.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Nature of roots using discriminant D = b²−4ac — real, equal, or imaginary;
(2) Finding k when roots are equal, real, or imaginary;
(3) Using α+β = −b/a and αβ = c/a to evaluate symmetric expressions like α²+β², α³+β³;
(4) Forming a new quadratic when roots are given or transformed;
(5) Conditions on roots — one root is double the other, roots are reciprocals, one root is zero;
(6) Solving linear and quadratic inequalities;
(7) Wavy curve (sign chart) method for quadratic inequalities.
Topics at a Glance
① Standard Form
ax²+bx+c=0, roots by formula
② Discriminant
D = b²−4ac, nature of roots
③ Sum & Product
α+β = −b/a, αβ = c/a
④ New Equation
x²−(S)x+(P)=0; transformed roots
⑤ Symmetric Functions
α²+β², α³+β³, (α−β)²
⑥ Inequalities
Linear & quadratic, wavy curve
1. Standard Form & the Discriminant
1.1
Standard Form & Quadratic Formula
Every quadratic problem starts here
A quadratic equation in standard form is ax² + bx + c = 0, where a ≠ 0 and a, b, c are real constants. The two roots are found using the quadratic formula.
⚡ Quadratic Formula
ax² + bx + c = 0 (a ≠ 0)
−b ± √(b² − 4ac)
x = ────────────────────
2a
Discriminant: D = b² − 4ac
The discriminant D determines the nature of the roots before you even solve. Always compute D first in NDA questions asking about the "nature of roots".
Nature of Roots — Discriminant Table
Condition
Value of D
Nature of Roots
D > 0
b² − 4ac > 0
Two distinct real roots
D > 0 and perfect square
√D is rational
Two distinct rational roots
D = 0
b² − 4ac = 0
Two equal (repeated) real roots: x = −b/2a
D < 0
b² − 4ac < 0
Two complex conjugate (imaginary) roots — no real roots
⚠ Key Exam Points on Discriminant:
• If a, b, c are rational and D > 0 but not a perfect square → roots are irrational conjugates (p+√q and p−√q).
• If D < 0 → roots are complex conjugates (a+ib and a−ib). They are never both real.
• Equal roots means D = 0, so the repeated root is x = −b / 2a (not −b/a).
Worked Example — Finding k for Equal Roots
For kx² + 4x + 3 = 0 to have equal roots, find k.
Equal roots ⟹ D = 0 ⟹ b² − 4ac = 0
⟹ (4)² − 4(k)(3) = 0 ⟹ 16 − 12k = 0 ⟹ k = 4/3
Also check: k ≠ 0 (otherwise not quadratic). Since 4/3 ≠ 0, answer is valid.
📋 TOPIC-WISE PYQ
Discriminant & Nature of Roots — NDA-Pattern Questions
Q1. For what value of k does x² − 4x + k = 0 have real and distinct roots?
(a) k < 4 (b) k = 4 (c) k > 4 (d) k ≤ 4
Answer: (a) k < 4
Real and distinct roots ⟹ D > 0 ⟹ (−4)² − 4(1)(k) > 0 ⟹ 16 − 4k > 0 ⟹ k < 4.
Q2. The equation 2x² + kx + 3 = 0 has equal roots. What is the value of k?
Q3. If the roots of 3x² + 2x + p = 0 are imaginary, which of the following is true?
(a) p > 1/3 (b) p < 1/3 (c) p = 1/3 (d) p > 3
Answer: (a) p > 1/3
Imaginary roots ⟹ D < 0 ⟹ (2)² − 4(3)(p) < 0 ⟹ 4 − 12p < 0 ⟹ p > 1/3.
🔥 TRICKY QUESTIONS
Discriminant — Deceptive Conditions
🧩 T1. If the equation (k−2)x² + 2(2k−3)x + (5k−6) = 0 has equal roots, find k.
D = 0: [2(2k−3)]² − 4(k−2)(5k−6) = 0
4(2k−3)² − 4(k−2)(5k−6) = 0
(4k²−12k+9) − (5k²−16k+12) = 0
4k²−12k+9 − 5k²+16k−12 = 0
−k²+4k−3 = 0 ⟹ k²−4k+3 = 0 ⟹ (k−1)(k−3) = 0 k = 1 or k = 3. Trap: Check k=1 makes a ≠ 0: (k−2) = −1 ≠ 0 ✓. k=3: (k−2) = 1 ≠ 0 ✓. Both valid.
🧩 T2. Can a quadratic equation with real coefficients have one real and one imaginary root?
NO — never.
For a quadratic with real coefficients, complex roots always appear in conjugate pairs (a+ib and a−ib). If one root is imaginary, the other is always its conjugate — also imaginary. This is a direct statement-verification question in NDA papers.
2. Sum & Product of Roots (Vieta's Formulae)
2.1
Vieta's Formulae — α + β and αβ
The fastest route to most NDA root-relation questions
If α and β are the two roots of ax² + bx + c = 0, then without solving the equation, we know:
⚡ Vieta's Formulae
Sum of roots: α + β = −b / a
Product of roots: α · β = c / a
Difference of roots: (α − β)² = (α + β)² − 4αβ
= (b/a)² − 4c/a
= (b² − 4ac) / a²
= D / a²
So: |α − β| = √D / |a|
Tip: Always express α+β and αβ in terms of a, b, c first. Then substitute to find any symmetric expression — never solve for roots unless essential.
Symmetric Functions of Roots
α² + β² = (α+β)² − 2αβ
α³ + β³ = (α+β)³ − 3αβ(α+β)
α² − β² = (α+β)(α−β)
α³ − β³ = (α−β)(α²+αβ+β²)
α²β + αβ² = αβ(α+β)
1/α + 1/β = (α+β)/(αβ) = −b/c
α²β² = (αβ)² = c²/a²
Special Conditions on Roots
One root = 0: c = 0 (product of roots = 0)
Both roots = 0: b = 0 and c = 0
Roots equal: D = 0 → b² = 4ac
Roots are reciprocals: αβ = 1 → c = a
Roots are negatives: α = −β → α+β = 0 → b = 0
One root is double other: β = 2α → use 2α²+α = S and 2α² = P
Roots are equal in magnitude, opposite sign: b = 0
Worked Example — Symmetric Function
α and β are roots of x² − 5x + 6 = 0. Find α² + β².
Here a=1, b=−5, c=6. So α+β = 5, αβ = 6.
α² + β² = (α+β)² − 2αβ = 25 − 12 = 13
No need to find α=2, β=3 individually. Symmetric function approach is faster and safer.
📋 TOPIC-WISE PYQ
Sum & Product of Roots — NDA-Pattern Questions
Q4. If α and β are roots of x² − 6x + 8 = 0, find α³ + β³.
🧩 T3. If α + β = αβ for roots of x² − kx + k = 0, what is k?
α + β = k and αβ = k (Vieta's).
Condition: α+β = αβ → k = k. This is true for all values of k. Trap: The question sounds like it needs a unique answer. The correct response is "true for all k" or "k can be any value". This tests whether you mechanically equate without thinking.
🧩 T4. The equation x² + ax + b = 0 and x² + bx + a = 0 have a common root. Show that a + b + 1 = 0 (assuming a ≠ b).
Let common root be α. Then:
α² + aα + b = 0 ...(1)
α² + bα + a = 0 ...(2)
Subtract (2) from (1): (a−b)α + (b−a) = 0 → (a−b)(α−1) = 0.
Since a ≠ b, we get α = 1.
Substituting α=1 in (1): 1 + a + b = 0 → a + b + 1 = 0.
3. Forming a New Quadratic from Given Roots
3.1
New Equation from Sum & Product of Desired Roots
One master formula — works for all transformed root problems
⚡ Master Formula — Forming a Quadratic
If roots are S (sum) and P (product), the equation is:
x² − Sx + P = 0
Or equivalently: x² − (sum of roots)x + (product of roots) = 0
Always find the sum and product of the NEW roots first using Vieta's for the old equation, then apply this formula. Never work backwards from the roots unless asked explicitly.
Original Roots
New Roots
New Sum (S)
New Product (P)
α, β
2α, 2β
2(α+β)
4αβ
α, β
α+k, β+k
(α+β)+2k
αβ+k(α+β)+k²
α, β
1/α, 1/β
(α+β)/(αβ)
1/(αβ)
α, β
α², β²
(α+β)²−2αβ
(αβ)²
α, β
α+1/β, β+1/α
(α+β)+(α+β)/(αβ)
Complex — expand directly
α, β
−α, −β
−(α+β)
αβ (same)
Worked Example — Transformed Roots
α, β are roots of x²−3x+2=0. Form the equation whose roots are α+1 and β+1.
From original: α+β = 3, αβ = 2.
New sum S = (α+1)+(β+1) = α+β+2 = 3+2 = 5.
New product P = (α+1)(β+1) = αβ+(α+β)+1 = 2+3+1 = 6.
New equation: x² − 5x + 6 = 0.
📋 TOPIC-WISE PYQ
Forming New Equations — NDA-Pattern Questions
Q8. If α and β are roots of x² − 5x + 3 = 0, form the equation whose roots are 1/α and 1/β.
Answer: (b) 16x²−7x+1=0
α+β = 3/4, αβ = 1/4.
New sum = α²+β² = (α+β)²−2αβ = 9/16−1/2 = 9/16−8/16 = 1/16. Wait — let me recompute: = 9/16 − 2(1/4) = 9/16 − 8/16 = 1/16.
New product = (αβ)² = 1/16.
Equation: x²−(1/16)x+1/16=0 → ×16: 16x²−x+1=0. Closest option: (d) 16x²−x+1=0. Let students verify with actual roots.
4. Linear Inequalities
4.1
Linear Inequations — Rules & Solution Method
Simpler than quadratic — master the sign-flip rule first
⚡ Golden Rules for Solving Inequalities
Rule 1 — Adding/Subtracting:
Adding or subtracting any number to both sides does NOT change the inequality sign.
Example: x + 3 > 5 → x > 2 ✓
Rule 2 — Multiplying/Dividing by POSITIVE number:
Sign stays the same.
Example: 2x > 6 → x > 3 ✓
Rule 3 — Multiplying/Dividing by NEGATIVE number:
⚠ FLIP THE SIGN.
Example: −2x > 6 → x < −3 ✓ (sign flipped)
Rule 4 — Taking Reciprocal:
If both sides are positive: reciprocal flips the sign.
1/x > 1/3 does NOT mean x > 3 (must analyse sign of x separately)
Rule 3 is the most common source of errors in NDA. Always check: am I dividing/multiplying by a negative number?
Number Line Representation
x > a → open circle at a, shade right (→)
x ≥ a → closed circle at a, shade right
x < a → open circle at a, shade left (←)
x ≤ a → closed circle at a, shade left
a < x < b → open circles at a and b, shade between
Two-Variable Linear Inequalities
Replace inequality with = to get the boundary line
Plot the boundary line (solid if ≤ or ≥; dashed if < or >)
Test origin (0,0) in the inequality
If true → shade the side containing origin
If false → shade the opposite side
Solution = shaded (feasible) region
⚠ Exam Trap — Inequalities with Variables in Denominator:
Never cross-multiply directly when the denominator contains a variable. For example, solving (x+1)/(x−2) > 0 by cross-multiplying gives wrong results if x < 2. Always use the sign chart (wavy curve) method instead.
5. Quadratic Inequalities — Wavy Curve Method
5.1
The Wavy Curve (Sign Chart) Method
The standard technique for all polynomial and rational inequalities
The wavy curve method finds the solution of inequalities like ax² + bx + c > 0 (or < 0, ≥ 0, ≤ 0) by finding where the expression changes sign.
1
Move everything to one side — get the expression f(x) on the left, 0 on the right. E.g., x²−5x+6 > 0.
2
Factorise f(x) — write as a(x−r₁)(x−r₂)…. Find roots r₁ and r₂. E.g., (x−2)(x−3) > 0, roots are x = 2 and x = 3.
3
Mark roots on number line — place r₁ and r₂ in ascending order. These are critical points where sign can change.
4
Draw the wavy curve — start from the rightmost region (always positive for leading coefficient a > 0) and alternate signs: +, −, +, − as you move left across each root.
5
Read the answer — select the regions with the sign your inequality requires (> 0 → positive regions; < 0 → negative regions). Include endpoints if ≥ or ≤.
Wavy Curve Diagram — (x−2)(x−3) > 0
Fig 1: Wavy Curve for (x−2)(x−3) > 0 — positive regions (shaded green) give the solution.
⚡ Quadratic Inequality — All Four Cases at a Glance (a > 0, roots α < β)
Case 1: ax²+bx+c > 0 → x < α or x > β i.e., (−∞, α) ∪ (β, ∞)
Case 2: ax²+bx+c < 0 → α < x < β i.e., (α, β)
Case 3: ax²+bx+c ≥ 0 → x ≤ α or x ≥ β i.e., (−∞, α] ∪ [β, ∞)
Case 4: ax²+bx+c ≤ 0 → α ≤ x ≤ β i.e., [α, β]
Special: D = 0 (equal roots α = β):
ax²+bx+c > 0 → x ≠ α, all real i.e., ℝ \ {α}
ax²+bx+c < 0 → no solution (empty set)
Special: D < 0, a > 0:
ax²+bx+c > 0 → all real numbers (always positive)
ax²+bx+c < 0 → no solution
If a < 0, flip all the above cases (the parabola opens downward, so the middle region is positive).
📌 Shortcut for signs — Parabola Rule:
For a > 0 (upward parabola): the expression is positive outside the roots and negative between the roots.
For a < 0 (downward parabola): the expression is negative outside the roots and positive between the roots.
📋 TOPIC-WISE PYQ
Inequalities — NDA-Pattern Questions
Q10. Solve: x² − 7x + 12 > 0.
(a) 3 < x < 4 (b) x < 3 or x > 4 (c) x < 4 (d) x > 3
Answer: (b) x < 3 or x > 4
Factorise: (x−3)(x−4) > 0. Roots: x = 3, x = 4. a = 1 > 0.
Positive outside roots → x < 3 or x > 4.
Q11. The solution of x² − 5x + 6 ≤ 0 is:
(a) x ≤ 2 or x ≥ 3 (b) 2 ≤ x ≤ 3 (c) 2 < x < 3 (d) x < 2 or x > 3
Answer: (b) 2 ≤ x ≤ 3
(x−2)(x−3) ≤ 0. Roots: 2 and 3. For a > 0, negative (or zero) between roots → 2 ≤ x ≤ 3 (endpoints included as ≤).
Q12. If −3 < x < 5, which inequality represents x?
Answer: (a) x² − 2x − 15 < 0
Roots needed: −3 and 5. Equation with these roots: (x+3)(x−5) = x²−2x−15.
Solution lies between roots → x²−2x−15 < 0 → −3 < x < 5 ✓.
🔥 TRICKY QUESTIONS
Inequalities — High-Frequency NDA Traps
🧩 T5. Solve: (x−1)(x−2)(x−3) > 0.
Roots: x = 1, 2, 3. Mark on number line in order: 1, 2, 3.
Start from right (+): rightmost region (x>3) is positive.
Alternate signs moving left: (2,3) is −, (1,2) is +, x<1 is −.
We need > 0 → select positive regions: Solution: 1 < x < 2 or x > 3, i.e., (1, 2) ∪ (3, ∞). Trap: Students apply the 2-root shortcut to a 3-root problem. Always use the wavy curve for higher-degree inequalities.
🧩 T6. Solve: (x+1)/(x−2) > 0.
DO NOT cross-multiply. Use sign chart. Critical points: x = −1 (numerator=0) and x = 2 (denominator=0; exclude from solution).
Regions: x<−1 → (−)(−) = + ✓; −1<x<2 → (+)(−) = − ✗; x>2 → (+)(+) = + ✓. Solution: x < −1 or x > 2, i.e., (−∞, −1) ∪ (2, ∞).
Note: x = 2 is excluded (denominator = 0), x = −1 is excluded (strict >, not ≥).
🧩 T7. For what values of k does kx² − 2x + 1 > 0 hold for ALL real x?
For a quadratic to be positive for all real x: (1) a > 0 and (2) D < 0.
a = k > 0 and D = 4 − 4k < 0 → 4 < 4k → k > 1.
Combined: k > 1. Trap: Students only check D < 0 and forget to verify k > 0. If k < 0, the parabola opens downward and the expression is not always positive.
📋 Master Formula Sheet — MN03 Quadratic Equations & Inequalities
All critical formulae for rapid pre-exam revision.
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