Quadratic Equations and Theory of Equations

Quadratic equations are among the most reliably tested Algebra topics in NDA — expect 3–5 questions across roots, nature of roots and forming equations. Most are one-step applications of the sum/product relations, so speed matters more than depth.

Core ideas

For ax² + bx + c = 0 (a ≠ 0), roots are x = [−b ± √(b²−4ac)] / 2a. The discriminant D = b²−4ac decides the nature:

• D > 0: two distinct real roots.
• D = 0: two equal (repeated) real roots.
• D < 0: two complex conjugate roots.

Sum and product of roots: if α, β are roots, α + β = −b/a and αβ = c/a. To build an equation from given roots: x² − (sum)x + (product) = 0.

Useful identities: α² + β² = (α+β)² − 2αβ; α − β = ±√[(α+β)² − 4αβ]; α³+β³ = (α+β)³ − 3αβ(α+β).

Roots are equal in magnitude opposite in sign when b = 0; reciprocal when a = c; one root zero when c = 0.

Exam Tricks & Tips

• 🎯 If a rational quadratic has one irrational root p+√q, the other MUST be p−√q (conjugate root theorem).
• 🎯 For complex coefficients this conjugate rule does not apply — check the coefficients are real first.
• 🎯 Product of roots negative ⇒ roots have opposite signs; product positive with sum positive ⇒ both positive.
• 🎯 To find the equation whose roots are 1/α, 1/β, replace x by 1/x in the original and clear denominators (reverses coefficients).
• 🎯 If both roots lie between numbers, check D≥0, af(k)>0, and the vertex location — a fast three-condition test.
• ❌ Common mistake: forgetting that D = 0 still gives real roots. 'Real and equal' is real; only D < 0 makes roots non-real.

Expected exam pattern

Items ask for nature of roots, value of an unknown k making roots equal/real, the sum α²+β², or forming a new quadratic with transformed roots. A common trap gives α+β and αβ and asks for a symmetric expression.

Quick recap

Lock in D = b²−4ac and its three cases, α+β = −b/a, αβ = c/a, x² −(sum)x +(product) = 0, and the symmetric-function identities. Irrational and complex roots come in conjugate pairs only for real coefficients.

Cover, recall, check. 11 cards on quadratic and root-relation results.

Q1. Quadratic formula for ax²+bx+c=0?
A1. x = [−b ± √(b²−4ac)] / 2a.

Q2. Discriminant and the three cases?
A2. D = b²−4ac. D>0 real distinct; D=0 real equal; D<0 complex conjugate.

Q3. Sum and product of roots?
A3. α+β = −b/a; αβ = c/a.

Q4. How to form a quadratic from roots α, β?
A4. x² − (α+β)x + αβ = 0.

Q5. α² + β² in terms of sum and product?
A5. (α+β)² − 2αβ.

Q6. α³ + β³ formula?
A6. (α+β)³ − 3αβ(α+β).

Q7. Conjugate root theorem?
A7. For real coefficients, if p+√q is a root so is p−√q; if a+ib is a root so is a−ib.

Q8. Condition for roots reciprocal to each other?
A8. a = c (so product αβ = 1).

Q9. Condition for roots equal magnitude, opposite sign?
A9. b = 0 (so sum = 0).

Q10. |α − β| in terms of coefficients?
A10. √D / |a| = √(b²−4ac)/|a|.

Q11. Equation with roots 1/α, 1/β from ax²+bx+c=0?
A11. cx² + bx + a = 0 (reverse the coefficients).

The quadratic ax² + bx + c = 0 is the most-tested single equation in NDA Mathematics — its roots, their nature, and the relations they obey appear every year.

Core idea / what this tests: Reading everything you need — sum, product, nature of roots — directly from the coefficients, without always solving.

Deep explanation

Beginner — Roots and the discriminant

The roots are x = [−b ± √(b² − 4ac)] / 2a. The discriminant D = b² − 4ac decides their nature:

• D > 0 → two distinct real roots.
• D = 0 → equal (repeated) real roots.
• D < 0 → complex conjugate roots.
  If a, b, c are rational and D is a perfect square, the roots are rational.

Intermediate — Sum and product of roots (Vieta)

For ax² + bx + c = 0 with roots α, β:

• α + β = −b/a
• αβ = c/a
  From these you build any symmetric function: α² + β² = (α+β)² − 2αβ; α − β = ±√D/a; 1/α + 1/β = (α+β)/(αβ).
  To form an equation from roots: x² − (sum)x + (product) = 0.

Advanced — Higher-degree equations

For a cubic ax³ + bx² + cx + d = 0 with roots α, β, γ:

• α + β + γ = −b/a
• αβ + βγ + γα = c/a
• αβγ = −d/a
  Complex and irrational roots occur in conjugate pairs when coefficients are real/rational. A polynomial of degree n has exactly n roots (counting multiplicity).

Worked example

Find k so that the roots of x² − kx + 9 = 0 are equal.
Equal roots need D = 0: (−k)² − 4·1·9 = 0 ⇒ k² = 36 ⇒ k = ±6.
Check: with k = 6, x² − 6x + 9 = (x−3)² = 0, giving the repeated root 3. ✓

NDA relevance

Typical asks: nature of roots for a parameter, value of a symmetric expression, forming an equation whose roots are related to a given one (e.g. reciprocals or shifted by a constant), and sign conditions ("both roots positive"). Each is a one-line Vieta application.

Exam tricks & shortcuts

• "Both roots positive" ⇒ sum > 0 and product > 0 and D ≥ 0.
• Roots reciprocal of the original ⇒ reverse the coefficients (a, b, c → c, b, a).
• To increase each root by h, replace x by (x − h).
• Mnemonic: Sum = −b/a, Product = +c/a ("minus b, plus c").

Key formulas

• ax² + bx + c = 0 ⇒ x = (−b ± √(b²−4ac))/2a.
• Sum of roots α+β = −b/a; product αβ = c/a.
• Discriminant D = b² − 4ac: D>0 real distinct, D=0 equal, D<0 complex.
• Equation from roots: x² − (α+β)x + αβ = 0.
• α² + β² = (α+β)² − 2αβ; |α−β| = √D/|a|.
• For real coefficients, complex roots occur in conjugate pairs.

Usage: use the sign of the discriminant to classify the roots before solving.

Worked Example

Problem: If α and β are the roots of x² − 6x + 10 = 0, form the quadratic equation whose roots are α² and β².

Solution:
The roots of x² − 6x + 10 = 0 are complex (discriminant = 36 − 40 = −4 < 0), so we work through Vieta's relations rather than by finding the roots.
From ax² + bx + c = 0: α + β = −b/a = 6 and αβ = c/a = 10.

We need the sum and product of the new roots α² and β².

Sum: α² + β² = (α + β)² − 2αβ = 6² − 2(10) = 36 − 20 = 16.

Product: α²β² = (αβ)² = 10² = 100.

A quadratic with roots p and q is x² − (p + q)x + pq = 0, so the required equation is
x² − 16x + 100 = 0.

Answer: x² − 16x + 100 = 0.