Algebra (General) for UPSC CDS — Equations, Identities, and Polynomials

Concept

Algebra at the CDS level is not about solving for x in a vacuum — it is about recognizing structure. The CDS paper consistently rewards candidates who can see a surd expression as a perfect square in disguise, or who know that a quadratic's roots telegraph the value of k without ever needing the quadratic formula.

Think of algebra as a grammar for numbers. Just as you do not decode a sentence word by word but read meaning in phrases, you should read algebraic expressions in chunks: recognizing a² - b² as (a+b)(a-b) in under a second, spotting that x = 7 + 4√3 almost certainly hides a squared binomial, seeing that "roots in ratio 2:3" immediately suggests substituting 2a and 3a.

Here is a useful mental model: every CDS algebra question has a locked door. The question gives you either the key (an identity, a theorem, a substitution) or a big enough hint that the key is nearby. Your job is to identify which key fits — not to brute-force the door with long calculations.

The major algebraic territories tested in CDS are:

• Surds and rationalization — simplifying √x, 1/√x, rationalizing denominators.
• Quadratic equations — discriminant, sum and product of roots (Vieta's formulas), roots in a given ratio.
• Polynomials — Remainder Theorem, Factor Theorem, synthetic evaluation.
• Cubic equations — Vieta's for three roots, recognizing standard factored forms.
• Indices and logarithms — chain equalities like 2ˣ = 3ʸ = 6ᶻ, change of base.
• Series — sum of first n odd numbers, A.P. and G.P. sum formulas.
• Discriminant conditions — for real/equal/complex roots.

None of these require anything beyond Class 10 theory. What CDS demands is fluency — getting from question to answer in 90 seconds, not 9 minutes.

Deep Dive

Surds and Perfect-Square Disguises

When you see x = a + b√c, your first instinct should be: can this be written as (p + q√c)²? Expanding (p + q√c)² = p² + q²c + 2pq√c. Match coefficients:

• Rational part: p² + q²c = a
• Irrational part: 2pq = b

For x = 7 + 4√3: rational part 7, irrational coefficient 4. Try p = 2, q = 1: p² + q²(3) = 4 + 3 = 7. Check 2pq = 2(2)(1) = 4. So x = (2 + √3)², meaning √x = 2 + √3 and 1/√x = 2 - √3. Their sum is 4. The whole question collapses in 20 seconds once you recognize the structure.

Vieta's Formulas — Your Most Reused Tool

For a quadratic ax² + bx + c = 0 with roots α and β:

$\alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a}$

For a cubic x³ + px² + qx + r = 0 with roots α, β, γ:

$\alpha + \beta + \gamma = -p, \quad \alpha\beta + \beta\gamma + \gamma\alpha = q, \quad \alpha\beta\gamma = -r$

Do not memorize these as isolated formulas — internalize the sign pattern: sum of roots = -(coefficient of second-highest term) / (leading coefficient). For x³ - 6x² + 11x - 6, the coefficient of x² is -6, so sum = -(-6)/1 = 6. That is the entire solution to one PYQ.

Roots in a Given Ratio

If roots are in ratio m:n, substitute α = ma and β = na. Then:

• ma + na = (m+n)a = -b/a → solve for a
• (ma)(na) = mn·a² = c/a → verify or find k

This technique bypasses the quadratic formula entirely. For ratio 2:3 and sum = 5: 5a = 5, so a = 1, roots are 2 and 3, product = 6.

The Remainder Theorem

When f(x) is divided by (x - a), the remainder is f(a). No long division needed. For f(x) = x³ - 3x² + 4x - 2 divided by (x - 1): substitute x = 1 directly: 1 - 3 + 4 - 2 = 0. Done in 8 seconds. If remainder is 0, then (x - 1) is a factor (Factor Theorem).

Discriminant Conditions

For ax² + bx + c = 0, discriminant D = b² - 4ac:

• D > 0: two distinct real roots
• D = 0: two equal real roots (perfect square)
• D < 0: no real roots

When a question gives a condition ("discriminant is zero"), set b² - 4ac = 0 and solve for the unknown parameter directly.

Chain Equalities with Indices and Logarithms

When aˣ = bʸ = cᶻ = k, take logarithms: x = log_a k, y = log_b k, z = log_c k. Equivalently, 1/x = log_k a, 1/y = log_k b, 1/z = log_k c. If c = ab, then log_k c = log_k a + log_k b, which gives 1/z = 1/x + 1/y. This is the key identity for any question of the form 2ˣ = 3ʸ = 6ᶻ.

Sum of First n Odd Numbers

The series 1 + 3 + 5 + ... + (2n-1) has sum n². This is not a derivation you need to redo every time — stamp it as a fact. The 50th odd number is 2(50) - 1 = 99, and 1 + 3 + ... + 99 = 50² = 2500.

G.P. Sum Formula

For a geometric progression with first term a, common ratio r ≠ 1, and n terms:

$S_n = \frac{a(r^n - 1)}{r - 1}$

For a = 1, r = 2: Sₙ = 2ⁿ - 1. Set equal to 255: 2ⁿ = 256 = 2⁸, so n = 8. Powers of 2 up to 2¹⁰ = 1024 should be memorized.

Memory Tricks & Shortcuts

Fast-Solving Framework

In the exam hall, classify the question in the first 5 seconds and apply the corresponding tool:

Surd / √x question → Try perfect-square factoring first. If x = a + b√c, attempt x = (p + q√c)².

Quadratic with condition on k → Discriminant = 0 (equal roots), discriminant > 0 (real roots), or roots in a ratio (use m:n substitution). Pick exactly one approach.

Polynomial divided by (x − a) → Plug x = a into f(x) directly. Do not divide.

Sum / product of roots asked → Apply Vieta's formulas. Never solve for individual roots unless the question explicitly asks for them.

Cubic root-sum question → Sum = −(coeff of x²), no factoring needed.

Chain equality aˣ = bʸ = cᶻ → Let all equal k, write reciprocals as logs, use log addition for product bases.

Series sum → Identify type: odd-number series (n²), AP (n/2 × (first + last)), GP (a(rⁿ − 1)/(r − 1)).

If none of the above fits cleanly in 10 seconds, use elimination: substitute answer choices into the original equation. CDS options are usually small integers.

Solved PYQs

Why this question: Tests whether you recognize a surd as a perfect square — the single most common surd trap in CDS.

Solving path: Write x = 7 + 4√3. Ask: is this (p + q√3)² for small integers? Try p = 2, q = 1: (2 + √3)² = 4 + 3 + 4√3 = 7 + 4√3. Yes. So √x = 2 + √3. Then 1/√x = 1/(2 + √3) = (2 - √3)/((2+√3)(2-√3)) = (2 - √3)/(4 - 3) = 2 - √3. Sum = (2 + √3) + (2 - √3) = 4.

Why this question: Classic Vieta's application with a ratio condition — tests whether you know the substitution trick or waste time with the quadratic formula.

Solving path: Roots in ratio 2:3 → write them as 2a and 3a. Sum of roots = 2a + 3a = 5a. From the equation x² - 5x + k = 0, sum = 5. So 5a = 5, a = 1. Roots are 2 and 3. Product of roots = 2 × 3 = 6 = k.

Why this question: Chain-equality with indices — CDS recycles this structure. You need the logarithm-reciprocal method, not guesswork.

Solving path: Let 2ˣ = 3ʸ = 6ᶻ = k. Then 1/x = log_k 2, 1/y = log_k 3, 1/z = log_k 6. Since 6 = 2 × 3: log_k 6 = log_k 2 + log_k 3, so 1/z = 1/x + 1/y. Check the options — option B matches exactly.

Why this question: Remainder Theorem application — if you know the theorem, this is a 10-second question. Without it, you spend 2 minutes on polynomial long division.

Solving path: Remainder Theorem: remainder when f(x) is divided by (x − 1) equals f(1). Compute f(1) = 1³ - 3(1²) + 4(1) - 2 = 1 - 3 + 4 - 2 = 0. Remainder is 0, so (x - 1) is a factor of the polynomial.

Why this question: Discriminant condition — standard template question. Tests whether you can set up b² - 4ac = 0 and isolate the parameter.

Solving path: Equation: kx² + 4x + 1 = 0. Here a = k, b = 4, c = 1. Discriminant = b² - 4ac = 16 - 4k. Set to zero: 16 - 4k = 0 → k = 4.

Common Mistakes

• Forgetting the sign in Vieta's sum formula. Sum of roots = −b/a, not b/a. In x² - 5x + k, b = −5, so sum = −(−5)/1 = 5. Many candidates write the equation as α + β = −5 and get the wrong value of a in ratio problems.

• Rationalizing √x instead of spotting the perfect square. When x = a + b√c, rationalization is slow and error-prone. Always check for the perfect-square form first. Rationalizing 1/√(7 + 4√3) without simplifying √x first adds three unnecessary steps.

• Applying the quadratic formula when Vieta's is enough. If a question asks only for the sum or product of roots, or for a parameter k given a root condition, Vieta's formulas give the answer directly. The quadratic formula finds individual roots — a task you rarely need.

• Wrong sign in cubic Vieta's. For x³ - 6x² + 11x - 6, the coefficient of x² is −6. Sum of roots = −(−6) = 6. A frequent error is reading the coefficient as +6 and writing the sum as −6.

• Using the wrong G.P. formula for the case r = 1. The formula Sₙ = a(rⁿ − 1)/(r − 1) is undefined at r = 1. For r = 1, Sₙ = na. This edge case rarely appears in CDS but do not auto-apply the standard formula without checking r.

• Confusing Remainder Theorem and Factor Theorem. Remainder Theorem: divide by (x − a), remainder = f(a). Factor Theorem: (x − a) is a factor if and only if f(a) = 0. These are related but distinct. In a question asking for the remainder, the answer may be non-zero — do not assume it is zero just because the polynomial looks "clean".

Related Topics

• quadratic equations
• progressions ap gp
• polynomials
• logarithms
• number systems
• indices surds

Practice on SarkariRise