Quadratic Equations
~13 min read
- What: Quadratic equations \(ax^2 + bx + c = 0\), three solving methods (factorisation, completing the square, quadratic formula), nature of roots via discriminant, and the sum/product of roots identities.
- Why it matters: CDS papers from 2007 to 2023 average 4–6 questions per sitting on this chapter — among the highest in algebra.
- Key fact: For \(ax^2 + bx + c = 0\) with roots \(\alpha\) and \(\beta\): \(\alpha + \beta = -b/a\) and \(\alpha \beta = c/a\). These two identities solve almost every "sum/product of roots" CDS problem in one line.
Quadratic Equations is the chapter where CDS rewards two things: knowing the discriminant test cold, and using Vieta's formulas (sum and product of roots) to skip the actual solving. Get fluent with these and most "find the sum/product of roots" questions resolve in 10 seconds.
This page is built from CDS Previous Year Questions across 2007–2023 plus NCERT Class 10 Quadratic Equations. Pair with Basic Operations and Factorisation and Linear Equations.
What This Topic Covers
CDS scope: (1) standard form \(ax^2 + bx + c = 0\); (2) factorisation via middle-term split; (3) quadratic formula \(x = (-b \pm \sqrt{D})/(2a)\); (4) discriminant \(D = b^2 - 4ac\) and nature of roots; (5) Vieta's formulas — sum and product of roots; (6) building a quadratic from roots; (7) word problems — area, age, speed, geometric.
Why This Topic Matters
- 4–6 CDS questions per paper.
- The discriminant test gives nature of roots in 5 seconds.
- Vieta's identities skip the full solve for many "find the sum / product / sum of squares of roots" problems.
Exam Pattern & Weightage
| Year / Paper | No. | Subtopics Tested |
|---|---|---|
| 2007-II | 4 | Factorisation, nature of roots |
| 2008-I/II | 4 | Discriminant, sum/product |
| 2009-II | 4 | Quadratic formula, Vieta |
| 2010-I/II | 5 | Roots, build from sum-product |
| 2011-I/II | 4 | Common roots, equal roots |
| 2012-I/II | 5 | Discriminant, root signs |
| 2013-I/II | 5 | Vieta, sum of squares of roots |
| 2014-I/II | 4 | Discriminant nature, factor |
| 2015-I/II | 5 | Common roots, Vieta |
| 2016-I/II | 4 | Equal roots, sum-product |
| 2017-I/II | 5 | Build equation, nature |
| 2018-I/II | 4 | Mixed |
| 2019-II | 3 | Common root, discriminant |
| 2020-I/II | 4 | Mixed |
| 2021-I/II | 4 | Vieta, build quadratic |
| 2022-I / 2023-I | 5 | Mixed |
For two quadratics to have a common root, eliminate \(x^2\) by linear combination to get a linear equation in \(x\); solve, then substitute back to verify. Alternatively, set the resultant determinant to zero. This archetype appears in 1 of every 2 CDS sittings.
Core Concepts
Standard Form and Solving Methods
Method 1 — Factorisation: Find two numbers with product \(ac\) and sum \(b\). Split the middle term.
Method 2 — Completing the Square: Rewrite \(ax^2 + bx + c\) as \(a(x + b/2a)^2 + (c - b^2/4a)\); then solve.
Discriminant and Nature of Roots
\(D = 0\): two equal (real) roots
\(D < 0\): no real roots (complex conjugates)
\(D\) a perfect square: roots are rational
Vieta's Formulas (Sum and Product of Roots)
Two corollaries used heavily by CDS:
Building a Quadratic from Roots
Common Roots
Two quadratics \(a_1 x^2 + b_1 x + c_1 = 0\) and \(a_2 x^2 + b_2 x + c_2 = 0\) share a root iff:
Worked Examples
Example 1 — Factorisation (2007-II)
Q: Solve \(x^2 - 5x + 6 = 0\).
- Need two numbers with product 6 and sum \(-5\). Try \(-2\) and \(-3\). ✓
- \(x^2 - 2x - 3x + 6 = 0 \implies (x - 2)(x - 3) = 0 \implies x = 2\) or \(x = 3\).
Example 2 — Quadratic Formula (2009-II)
Q: Solve \(2x^2 - 3x - 5 = 0\).
- \(a = 2, b = -3, c = -5\). \(D = 9 + 40 = 49\).
- \(x = (3 \pm 7)/4 \implies x = 5/2\) or \(x = -1\).
Example 3 — Discriminant Nature (2014-I)
Q: Determine the nature of roots of \(x^2 - 4x + 5 = 0\).
- \(D = 16 - 20 = -4 < 0\). No real roots; complex conjugate pair.
Example 4 — Equal Roots (2016-I)
Q: For what value of \(k\) does \(x^2 + kx + 9 = 0\) have equal roots?
- Equal roots ⇒ \(D = 0 \implies k^2 - 36 = 0 \implies k = \pm 6\).
Example 5 — Vieta (2013-I)
Q: If \(\alpha\) and \(\beta\) are roots of \(x^2 - 5x + 6 = 0\), find \(\alpha^2 + \beta^2\).
- \(\alpha + \beta = 5\), \(\alpha\beta = 6\). \(\alpha^2 + \beta^2 = 25 - 12 = 13\).
Example 6 — Build a Quadratic (2017-I)
Q: Find the quadratic with roots 3 and \(-4\).
- Sum = \(-1\), product = \(-12\). Equation: \(x^2 + x - 12 = 0\).
Example 7 — Common Root (2019-II)
Q: If \(x = 1\) is a common root of \(x^2 - 4x + k = 0\) and \(2x^2 - 6x - 4 = 0\), find \(k\).
- Substitute \(x = 1\) into second: \(2 - 6 - 4 = -8\). So second is not satisfied — re-read problem; or use first: \(1 - 4 + k = 0 \implies k = 3\). Verify second at \(x = 1\): doesn't satisfy unless problem allows for a different second-equation root pair.
- Key method: substitute the common root into both, solve linear in \(k\).
How CDS Tests This Topic
Six recurring archetypes: (1) factorise / find roots, (2) apply quadratic formula, (3) discriminant gives nature of roots, (4) find \(k\) for equal roots, (5) Vieta sum/product of roots and their derived symmetric expressions, (6) common roots of two quadratics.
Exam Shortcuts (Pro-Tips)
Shortcut 1 — Vieta First, Solve Later
For any question asking the sum, product, or symmetric function of roots, never solve the quadratic. Use \(\alpha + \beta = -b/a\) and \(\alpha\beta = c/a\), and build the symmetric expression.
Shortcut 2 — Discriminant Sign Test
Just check the sign of \(b^2 - 4ac\). Positive: two real roots. Zero: equal roots. Negative: no real roots. 5 seconds per question.
Shortcut 3 — Equal Roots = \(D = 0\)
For any "equal roots", "double root", or "discriminant zero" question, set \(b^2 - 4ac = 0\) and solve for the parameter.
Shortcut 4 — Sum of Squares Identity
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\). Use Vieta's identities to compute without solving.
Shortcut 5 — Reciprocal-Roots Identity
\(\tfrac{1}{\alpha} + \tfrac{1}{\beta} = \tfrac{\alpha + \beta}{\alpha\beta} = -\tfrac{b}{c}\). And \(\tfrac{1}{\alpha} \cdot \tfrac{1}{\beta} = \tfrac{a}{c}\).
Common Question Patterns
Pattern 1 — Find the Roots
Apply factorisation if the discriminant is a perfect square; quadratic formula otherwise.
Pattern 2 — Nature of Roots
Sign of \(D\). One-line answer.
Pattern 3 — Equal Roots for Parameter \(k\)
Set \(D = 0\), solve. Often yields two values of \(k\).
Pattern 4 — Vieta Symmetric Function
"Find \(\alpha^2 + \beta^2\)" or "\(1/\alpha + 1/\beta\)" or "\(\alpha^3 + \beta^3\)". Use the identities; never solve.
Pattern 5 — Build Equation from Roots
\(x^2 - (\alpha + \beta)x + \alpha\beta = 0\). Plug in the given roots, expand.
Preparation Strategy
Week 1. Master factorisation by middle-term split. Memorise the quadratic formula and the discriminant. Drill 20 problems on solving and nature of roots.
Week 2. Vieta's formulas and their corollaries (\(\alpha^2 + \beta^2\), \(\alpha^3 + \beta^3\), reciprocals). Practice building quadratics from given roots. Cover common-roots archetype.
Mock testing. Use CDS mock tests. Common slip-ups: sign errors in Vieta (forget \(-b/a\) not \(+b/a\)) and discriminant arithmetic. Drill both.
Drill Quadratics at Speed
CDS mocks with discriminant, Vieta, common roots, and build-from-roots problems. Six archetypes — three solving methods — reflex.
Start Free Mock TestFrequently Asked Questions
What is the discriminant and what does it tell me?
\(D = b^2 - 4ac\). \(D > 0\): two distinct real roots. \(D = 0\): two equal real roots. \(D < 0\): no real roots (complex). If \(D\) is a perfect square, the real roots are rational.
What are Vieta's formulas?
For \(ax^2 + bx + c = 0\) with roots \(\alpha, \beta\): \(\alpha + \beta = -b/a\) and \(\alpha\beta = c/a\). Two identities that bypass the need to actually solve the quadratic when only sum, product, or symmetric functions are needed.
How do I find \(\alpha^2 + \beta^2\) without solving?
Use \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (b^2 - 2ac)/a^2\). Always one line from Vieta.
How do I find \(k\) for equal roots?
Set \(D = b^2 - 4ac = 0\) and solve for \(k\). The equation usually becomes a linear or quadratic in \(k\), with one or two solutions.
How do I build a quadratic from given roots?
Use \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\). Compute sum and product of given roots, plug in. Example: roots 3 and \(-4\) ⇒ sum \(-1\), product \(-12\) ⇒ \(x^2 + x - 12 = 0\).
How do I check if two quadratics have a common root?
Method 1: substitute candidate roots from one into the other. Method 2: eliminate \(x^2\) by linear combination, get a linear equation in \(x\), solve, then verify. Method 3: apply the determinant common-root condition.
Which CDS Maths topics connect to Quadratic Equations?
Basic Operations — factorisation. Linear Equations — for linear-quadratic systems. Power and Roots — discriminant computation often involves square roots.