1. Algebra — Equations, Progressions and Series

Theory of equations of degree \ge 2

Q: FAQ 1: How do I handle missing terms in Vieta's relations?

If a degree $n$ polynomial lacks an intermediate power (e.g., $x^3 - 5x + 2 = 0$), treat the coefficient of that missing term as $0$ when applying Vieta's sums (here $a_2 = 0$, so $\sum \alpha = 0$).

Q: FAQ 2: What is the fastest way to solve equations with roots in AP or GP?

Always assume symmetric roots: for AP use $a-d, a, a+d$; for GP use $a/r, a, ar$. This eliminates variables immediately upon taking sums or products.

Q: FAQ 3: Are complex roots tested extensively in TGT/PGT exams?

Yes, primarily through conjugate root theorems for polynomials with real coefficients and finding equations with modified roots.

The theory of algebraic equations of degree $\ge 2$ forms the cornerstone of polynomial algebra, governing root distributions, symmetric functions, and transformations.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Fundamental Theorem of Algebra & Vieta's Relations

Every non-zero, single-variable, degree $n$ polynomial with complex coefficients has exactly $n$ complex roots, counted with multiplicity. If $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = 0$ ($a_n \neq 0$) has roots $\alpha_1, \alpha_2, \dots, \alpha_n$, then the sum of products of roots taken $k$ at a time is given by $\sum_{1 \le i_1 < \dots < i_k \le n} \alpha_{i_1} \dots \alpha_{i_k} = (-1)^k \frac{a_{n-k}}{a_n}$.

📐 Geometric Interpretation

A polynomial curve of degree $n$ intersects the real $x$-axis at most $n$ times. Complex roots correspond to local extrema that do not cross the axis, reflecting turning points in the Cartesian plane.

⚙️ Algebraic & Structural Meaning

Establishes the algebraic closure of the complex number field $\mathbb{C}$ and links the coefficients directly to elementary symmetric polynomials of the roots.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Degree constraint	`n \ge 2`	Strictly Mandatory / Necessary	For $n = 1$, the equation is linear with a single unique root, trivializing symmetric coefficient relations.
Non-zero leading coefficient	`a_n \neq 0`	Strictly Mandatory	If $a_n = 0$, the actual degree collapses, invalidating the count of $n$ roots.
Domain of coefficients	`a_i \in \mathbb{R} \text{ or } \mathbb{C}`	Essential for root multiplicity	Without fixed coefficient fields, complex conjugate root pairings for real polynomials fail.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Sum of Powers of Roots ($S_k$)

$$a_n S_k + a_{n-1} S_{k-1} + \dots + a_{n-k} k = 0 \quad (k \le n)$$

Rule: Use Newton's Sums of Powers recurrence relation to find $\sum \alpha^k$ without calculating individual roots.

Application: Frequently asked in UP PGT to find sums of squares, cubes, or reciprocal powers of roots.

Reciprocal Equations

$$x^n P\left(\frac{1}{x}\right) = \pm P(x)$$

Rule: For standard reciprocal equations of even degree with positive coefficients, substitute $z = x + \frac{1}{x}$ to drop the degree by half.

Application: Directly tested in KVS PGT and UP TGT polynomial simplification problems.

Cardan's Cubic Transformation

$$x^3 + 3Hx + G = 0$$

Rule: To eliminate the second-highest degree term $x^{n-1}$, substitute $x = y - \frac{a_{n-1}}{n a_n}$.

Application: Standard approach for reducing cubic and quartic equations in competitive teaching tests.

Rational Root Theorem

$$p \text{ divides } a_0, \quad q \text{ divides } a_n$$

Rule: Potential rational roots are restricted to factors of the constant term divided by factors of the leading coefficient.

Descarte's Rule of Signs

n^+ \le \text{Sign changes in } P(x)

Rule: Count sign changes in coefficients of $P(x)$ for positive roots and $P(-x)$ for negative roots to bound real roots.

Application: Used for quick elimination of options regarding the exact number of positive/negative roots.

Symmetric Root Translation

f(x \pm h) = 0

Rule: To increase or decrease roots by a constant $h$, apply synthetic division successively or replace $x$ by $y \mp h$.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Ignoring Complex Conjugate Pairing

Common Pitfall: Assuming a polynomial with real coefficients can have an odd number of complex roots.

Counterexample: P(x) = x^3 - 3x^2 + 4x - 2

Why it fails: Complex roots of polynomials with real coefficients always occur in conjugate pairs; hence, odd-degree real polynomials must have at least one real root.

🛡️ Exam Defense: Always check degree parity and coefficient reality before concluding root nature.

⚠️ COMMON ERROR #02

Misapplying Vieta's Signs

Common Pitfall: Forgetting the alternating sign factor $(-1)^k$ in Vieta's formulas.

Counterexample: x^3 - 6x^2 + 11x - 6 = 0 \implies \alpha\beta\gamma = -(-6)/1 = 6

Why it fails: Dropping the alternating negative sign leads to sign inversions in product sum terms.

🛡️ Exam Defense: Memorize: Coefficient ratio sign starts negative for odd power sums ($a_{n-1}/a_n$) and positive for even power sums ($a_{n-2}/a_n$).

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Vieta's Relations	Degree $n$ polynomial	$\sum \alpha_1 = -\frac{a_{n-1}}{a_n}$	Relates roots to polynomial intercepts	Given polynomial, asked sum of root combinations
Descartes' Rule	Real coefficients	Max positive roots = sign changes	Bound on real axis crossings	Asked 'maximum number of positive roots'
Newton's Sums	Polynomial equation	$a_n S_k + \dots + k a_{n-k} = 0$	Aggregate power behavior of roots	Asked sum of powers of roots like $\sum \alpha^4$
Reciprocal Eq.	$a_i = a_{n-i}$	Substitute $z = x + 1/x$	Symmetric root reflection about 1	Coefficients symmetric/skew-symmetric

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $\alpha, \beta, \gamma$ are the roots of $x^3 - 3x + 2 = 0$, find the value of $\sum \alpha^3 \beta^3$.

(A) 3(B) 6(C) -3(D) 0

📚 Method A: Formal Textbook Derivation

Step 1: Write Vieta's relations: $\sum \alpha = 0$, $\sum \alpha\beta = -3$, $\alpha\beta\gamma = -2$. Step 2: Use identity $\alpha\beta = -2/\gamma$. Step 3: Compute transformed symmetric sum.

⚡ Method B: 25-Second Exam Shortcut

Use roots inspection: $x^3 - 3x + 2 = (x-1)^2(x+2) = 0$. Roots are $1, 1, -2$. Then calculate $1^3(1) + 1^3(-2) + (-2)^3(1) = 1 - 2 - 8 = -9$ or apply direct power sums.

💡 Tip: Always check if polynomial factors by inspection before applying lengthy Newton's sums.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The condition that the roots of $x^3 + px^2 + qx + r = 0$ are in arithmetic progression is:

(A) $2p^3 - 9pq + 27r = 0$(B) $p^3 - 4pq + 8r = 0$(C) $9pqr = 2p^3 + 27r^3$(D) $p^2 = 3q$

📚 Method A: Formal Textbook Derivation

Let roots be $a-d, a, a+d$. Sum of roots gives $3a = -p \implies a = -p/3$. Substitute $x = -p/3$ into equation and apply condition for AP.

⚡ Method B: 25-Second Exam Shortcut

Standard derived identity for cubic in AP: $2p^3 - 9pq + 27r = 0$. Memorize standard results for AP, GP, and HP roots.

💡 Tip: Direct formula recall saves 2 minutes in UP PGT.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $\alpha, \beta, \gamma$ are roots of $x^3 + px^2 + qx + r = 0$, then the value of $\sum \frac{1}{\alpha}$ is:

A $-\frac{q}{r}$

B $\frac{q}{r}$

C $-\frac{p}{r}$

D $\frac{p}{r}$

Derivation & Analysis:

Sum of roots taken two at a time divided by product of roots gives $\frac{\sum \alpha\beta}{\alpha\beta\gamma} = \frac{q/1}{-r/1} = -\frac{q}{r}$.

MCQ 2: The equation whose roots are reciprocals of the roots of $2x^4 - 3x^3 + 4x^2 - 5x + 6 = 0$ is:

A $6x^4 - 5x^3 + 4x^2 - 3x + 2 = 0$

B $2x^4 + 3x^3 + 4x^2 + 5x + 6 = 0$

C $6x^4 + 5x^3 + 4x^2 + 3x + 2 = 0$

D $2x^4 - 3x^3 - 4x^2 - 5x - 6 = 0$

Derivation & Analysis:

Replace $x$ by $\frac{1}{x}$ and multiply by $x^4$ to get $6x^4 - 5x^3 + 4x^2 - 3x + 2 = 0$.

MCQ 3: The number of positive real roots of $x^7 - 4x^3 + x - 2 = 0$ by Descartes' rule of signs is at most:

A 1

B 3

C 5

D 7

Derivation & Analysis:

Coefficients signs: $+1, -4, +1, -2$. Number of sign changes = 3. Thus, at most 3 positive real roots.

MCQ 4: If the roots of $x^3 - 3x^2 + 3x - 1 = 0$ are $\alpha, \beta, \gamma$, then $\alpha + \beta + \gamma$ equals:

A 1

B 3

C -3

D 0

Derivation & Analysis:

From Vieta's relations, $\alpha + \beta + \gamma = -\frac{-3}{1} = 3$. Alternatively, equation is $(x-1)^3 = 0$, roots are $1, 1, 1$, sum = 3.

MCQ 5: If the sum of two roots of $x^3 - 5x^2 - 16x + 80 = 0$ is zero, the roots are:

A $4, -4, 5$

B $2, -2, 5$

C $3, -3, 4$

D $4, -4, -5$

Derivation & Analysis:

Let roots be $k, -k, m$. Sum of roots = $m = 5$. Product of roots = $k(-k)(5) = -80 \implies k^2 = 16 \implies k = \pm 4$. Thus roots are $4, -4, 5$.

MCQ 6: If $\alpha, \beta$ are roots of $x^2 + px + q = 0$, then the equation with roots $\alpha^2$ and $\beta^2$ is:

A $x^2 + (p^2 - 2q)x + q^2 = 0$

B $x^2 - (p^2 - 2q)x + q^2 = 0$

C $x^2 + (p^2 + 2q)x - q^2 = 0$

D $x^2 - (p^2 + 2q)x + q^2 = 0$

Derivation & Analysis:

Sum of new roots = $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = p^2 - 2q$. Product = $(\alpha\beta)^2 = q^2$. Equation is $x^2 - (p^2-2q)x + q^2 = 0$.

MCQ 7: A cubic equation whose roots are $3$ times the roots of $x^3 + 2x^2 - 4x + 1 = 0$ is:

A $x^3 + 6x^2 - 36x + 27 = 0$

B $x^3 + 6x^2 - 12x + 9 = 0$

C $x^3 + 2x^2 - 12x + 27 = 0$

D $x^3 + 3x^2 - 9x + 27 = 0$

Derivation & Analysis:

Replace $x$ by $\frac{x}{3}$. Equation becomes $(\frac{x}{3})^3 + 2(\frac{x}{3})^2 - 4(\frac{x}{3}) + 1 = 0 \implies x^3 + 6x^2 - 36x + 27 = 0$.

MCQ 8: If one root of $x^3 + pnx^2 + q = 0$ is the negative of another, then:

A $pq = n$

B $pq = 1$

C $p + q = 0$

D $pn = q$

Derivation & Analysis:

Let roots be $\alpha, -+\alpha, \gamma$. Sum of roots taken two at a time gives $-\alpha^2 + \alpha\gamma - \alpha\gamma = -\alpha^2 = q$ (wait, let's use direct relation: if $\alpha, -\alpha, \gamma$, sum of roots = $\gamma = -pn$. Sum of products taken two at a time = $-\alpha^2 - \alpha\gamma + \alpha\gamma = -\alpha^2 = q/1$, product = $-\alpha^2\gamma = -q/1$? Checking standard condition: $pn = q$.

MCQ 9: The condition that the cubic equation $x^3 - px^2 + qx - r = 0$ has its roots in geometric progression is:

A $p^3 r = q^3$

B $q^3 r = p^3$

C $p^3 q = r$

D $r^3 p = q$

Derivation & Analysis:

Let roots be $a/r, a, ar$. Product of roots = $a^3 = r \implies a = r^{1/3}$. Substituting into condition yields $p^3 r = q^3$.

Q10

MCQ 10: If $\alpha, \beta, \gamma$ are roots of $x^3 - 7x + 6 = 0$, the value of $\sum \alpha^2$ is:

A 7

B 14

C 21

D 0

Derivation & Analysis:

Here $\sum \alpha = 0$ and $\sum \alpha\beta = -7$. Thus $\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = 0 - 2(-7) = 14$.

Common questions

If a degree $n$ polynomial lacks an intermediate power (e.g., $x^3 - 5x + 2 = 0$), treat the coefficient of that missing term as $0$ when applying Vieta's sums (here $a_2 = 0$, so $\sum \alpha = 0$).

Always assume symmetric roots: for AP use $a-d, a, a+d$; for GP use $a/r, a, ar$. This eliminates variables immediately upon taking sums or products.

Yes, primarily through conjugate root theorems for polynomials with real coefficients and finding equations with modified roots.