1. Algebra — Equations, Progressions and Series

Quadratic equation

Q: FAQ 1: Why do students frequently misapply the condition for common roots in UP PGT exams?

Candidates often memorize the cross-multiplication determinant formula without first checking if the two quadratic equations are proportional. If $a_1/a_2 = b_1/b_2 = c_1/c_2$, both roots are common, rendering the standard single-root determinant identity inapplicable.

Q: FAQ 2: What is the most common reason aspirants lose marks in quadratic inequality questions?

Neglecting the sign of the leading coefficient $a$ when opening intervals or determining ranges. Always ensure $a > 0$ before applying standard interior/exterior interval rules for quadratic inequalities.

A quadratic equation is a second-order polynomial equation in a single variable, expressed canonically as $ax^2 + bx + c = 0$ with $a \neq 0$.

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

📖 Topic progress

○ Mark as complete

Definition and statement

Fundamental Theorem of Quadratic Equations and Discriminant Criterion

Let $ax^2 + bx + c = 0$ be a quadratic equation with real coefficients $a, b, c \in \mathbb{R}$ and $a \neq 0$. The roots are given explicitly by the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. The discriminant $\Delta = b^2 - 4ac$ completely determines the nature of the roots in $\mathbb{R}$ and $\mathbb{C}$.

📐 Geometric Interpretation

Geometrically, $y = ax^2 + bx + c$ represents a parabola whose axis of symmetry is $x = -\frac{b}{2a}$ and vertex is $\left(-\frac{b}{2a}, -\frac{\Delta}{4a}\right)$. The roots of the quadratic equation represent the exact $x$-intercepts (real zeros) where the parabola intersects the horizontal $x$-axis.

⚙️ Algebraic & Structural Meaning

Algebraically, it exemplifies the factorization of polynomials over $\mathbb{R}$ and $\mathbb{C}$. By Vieta's relations, the sum of roots $\alpha + \beta = -\frac{b}{a}$ and product of roots $\alpha\beta = \frac{c}{a}$ link polynomial coefficients directly to symmetric functions of roots.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Non-zero leading coefficient ($a \neq 0$)	`Algebraic constraint`	Strictly Mandatory / Necessary	If $a = 0$, the equation degrades into a linear equation $bx + c = 0$, destroying the quadratic parabola and root multiplicity properties.
Real coefficients for discriminant sign tests ($a, b, c \in \mathbb{R}$)	`Coefficient domain`	Strictly Mandatory	If coefficients are complex, the standard discriminant rules ($\Delta > 0 \implies$ real distinct roots) fail because complex numbers lack a standard total ordering.
Universal domain of roots ($x \in \mathbb{C}$)	`Solution space`	Essential for completeness	If restricted strictly to $\mathbb{R}$ when $\Delta < 0$, the equation possesses no solutions, violating the Fundamental Theorem of Algebra.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Location of Roots via Function Values

$$af(k) < 0 \implies \text{both roots lie on opposite sides of } k$$

Rule: To check if a given real number $k$ lies between the roots of $ax^2 + bx + c = 0$ (with $a > 0$), simply evaluate $f(k)$. If $f(k) < 0$, then $k$ is strictly between the roots without solving for them.

Application: Frequently asked in UP PGT to find the range of parameters $p$ for which roots bracket a specific integer.

Common Root Condition

$$ (a_1 b_2 - a_2 b_1)(b_1 c_2 - b_2 c_1) = (c_1 a_2 - c_2 a_1)^2 $$

Rule: For two quadratic equations $a_1 x^2 + b_1 x + c_1 = 0$ and $a_2 x^2 + b_2 x + c_2 = 0$ to have exactly one common root, use this determinant cross-multiplication identity instantly.

Application: Standard recurring pattern in DSSSB PGT and UP TGT mathematics.

Maximum and Minimum Values of Quadratic Expressions

$$ \text{Extremum value} = \frac{4ac - b^2}{4a} at x = -\frac{b}{2a} $$

Rule: If $a > 0$, the expression has a global minimum at the vertex. If $a < 0$, it has a global maximum. Never waste time completing the square.

Equations Reducible to Quadratics (Reciprocal Equations)

$$ \text{Substitute } z = x + \frac{1}{x} \text{ for type } ax^4 + bx^3 + cx^2 + bx + a = 0 $$

Rule: For symmetric coefficient reciprocal equations of even degree, divide by $x^2$ and substitute $z = x + \frac{1}{x}$ to reduce a quartic equation into a simple quadratic in $z$.

Range of Rational Expressions

$$ \text{Let } y = \frac{ax^2 + bx + c}{dx^2 + ex + f} \implies (ay-d)x^2 + (by-e)x + (cy-f) = 0 \text{ and set } \Delta \ge 0 $$

Rule: To find the range of a rational quadratic expression where $x \in \mathbb{R}$, cross-multiply, form a quadratic in $x$, and apply the reality condition $\Delta \ge 0$ to get an inequality in $y$.

Rapid Sign Inspection via Vieta's Sum and Product

$$ \text{If } \alpha\beta > 0 \text{ and } \alpha+\beta > 0 \implies \text{both positive; If } \alpha\beta < 0 \implies \text{opposite signs} $$

Rule: Instead of computing $\Delta$ and exact roots, check the sign of the product $c/a$ and sum $-b/a$ to instantly eliminate wrong options regarding root signs.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Forgetting Leading Coefficient Condition ($a \neq 0$)

Common Pitfall: Candidates blindly apply discriminant formulas to equations given as $(a-2)x^2 + 3x + 5 = 0$ without testing $a = 2$.

Counterexample: Equation $(m-1)x^2 + 2mx + 1 = 0$ claimed to be quadratic for all $m \in \mathbb{R}$.

Why it fails: If $m = 1$, the $x^2$ term vanishes, turning the equation linear ($2x + 1 = 0$), which yields only one root instead of two.

🛡️ Exam Defense: Always check if the coefficient of $x^2$ contains a parameter, and examine what happens when that coefficient equals zero.

⚠️ COMMON ERROR #02

Both Common Roots vs. Single Common Root Confusion

Common Pitfall: Applying the single common root condition when both roots are identical.

Counterexample: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

Why it fails: If both roots are common, the equations are proportional. Using the cross-product condition alone without verifying proportionality can lead to extraneous parameter values.

🛡️ Exam Defense: Always test proportionality $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ first when dealing with common root problems.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Nature of Roots	$a, b, c \in \mathbb{R}, a \neq 0$	$\Delta = b^2 - 4ac$	Parabola intersection with $x$-axis	Discriminant questions
Vieta's Relations	$ax^2 + bx + c = 0$	Sum $= -\frac{b}{a}$, Prod $= \frac{c}{a}$	Symmetric root properties	Symmetric polynomial evaluation
Location of Roots	$a > 0, f(k)$ evaluated	$af(k) < 0$ or condition set	Parabola sign relative to point $k$	Parameter range for root positions
Common Root	Two quadratic equations	$(a_1b_2 - a_2b_1)(b_1c_2 - b_2c_1) = (c_1a_2 - c_2a_1)^2$	Intersection of two parabolas	Find condition for shared solution

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If the roots of the equation $x^2 - px + q = 0$ differ by 1, then:

(A) $p^2 = 4q + 1$(B) $p^2 = 4q - 1$(C) $q^2 = 4p + 1$(D) $p^2 = q^2 + 4$

📚 Method A: Formal Textbook Derivation

Let roots be $\alpha$ and $\alpha + 1$. Sum of roots $\alpha + (\alpha + 1) = p \implies 2\alpha + 1 = p \implies \alpha = \frac{p-1}{2}$. Product of roots $\alpha(\alpha+1) = q$. Substitute $\alpha$: $\left(\frac{p-1}{2}\right)\left(/frac{p+1}{2}\right) = q \implies \frac{p^2 - 1}{4} = q \implies p^2 - 1 = 4q \implies p^2 = 4q + 1$.

⚡ Method B: 25-Second Exam Shortcut

Use $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$. Here $\alpha - \beta = 1$, so $1^2 = p^2 - 4q \implies p^2 = 4q + 1$. Done in 10 seconds.

💡 Tip: Whenever root difference is given, immediately invoke $(\alpha - \beta)^2 = (\sum\alpha)^2 - 4\prod\alpha$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If $x$ is real, the maximum value of the expression $\frac{1}{3x^2 - 6x + 7}$ is:

(A) $\frac{1}{4}$(B) $\frac{1}{3}$(C) $4$(D) $\frac{1}{2}$

📚 Method A: Formal Textbook Derivation

A rational expression is maximized when its denominator is minimized. The denominator is a quadratic $f(x) = 3x^2 - 6x + 7$. Its minimum value occurs at $x = -\frac{b}{2a} = -\frac{-6}{2(3)} = 1$. Minimum value = $\frac{4(3)(7) - (-6)^2}{4(3)} = \frac{84 - 36}{12} = \frac{48}{12} = 4$. Thus, maximum value of the fraction is $\frac{1}{4}$? Wait, let's recompute vertex: $f(1) = 3(1)^2 - 6(1) + 7 = 4$. So min denominator is 4, max value of fraction is $1/4$. Let's check options: (A) 1/4.

⚡ Method B: 25-Second Exam Shortcut

Direct vertex evaluation: denominator min at $x = 1$ is $3(1) - 6 + 7 = 4$. Reciprocal is $1/4$.

💡 Tip: Remember: max of reciprocal is reciprocal of minimum.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If the roots of $x^2 + px + q = 0$ are in the ratio $2:3$, then:

A $6p^2 = 25q$

B $p^2 = 25q$

C $6q^2 = 25p$

D $25p^2 = 6q$

Derivation & Analysis:

Let roots be $2\alpha$ and $3\alpha$. Sum $= 5\alpha = -p$, Product $= 6\alpha^2 = q$. Eliminating $\alpha$: $\alpha = -p/5 \implies 6(-p/5)^2 = q \implies 6p^2 / 25 = q \implies 6p^2 = 25q$.

MCQ 2: The value of $m$ for which the equation $mx^2 + 2x + m = 0$ has equal roots is:

A $m = \pm 1$

B $m = 0$

C $m = 1$

D $m = -1$

Derivation & Analysis:

For equal roots, $\Delta = 0 \implies b^2 - 4ac = 0 \implies 2^2 - 4(m)(m) = 0 \implies 4 - 4m^2 = 0 \implies m^2 = 1 \implies m = \pm 1$. Since $m \neq 0$, both $+1$ and $-1$ work.

MCQ 3: If $\alpha, \beta$ are roots of $x^2 - 5x + 6 = 0$, then the equation whose roots are $\alpha+3$ and $\beta+3$ is:

A $x^2 - 11x + 30 = 0$

B $x^2 - 8x + 20 = 0$

C $x^2 - 11x + 20 = 0$

D $x^2 + 11x + 30 = 0$

Derivation & Analysis:

Replace $x$ with $x-3$ in the original equation: $(x-3)^2 - 5(x-3) + 6 = 0 \implies x^2 - 6x + 9 - 5x + 15 + 6 = 0 \implies x^2 - 11x + 30 = 0$.

MCQ 4: If $a, b, c$ are in AP, then the roots of $ax^2 + bx + c = 0$ are real and always include:

A $1$

B $-1$

C $2$

D $0$

Derivation & Analysis:

Since $a, b, c$ are in AP, $2b = a + c$. Substitute $b = (a+c)/2$ into $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ or test $x = 1$: $a(1)^2 + b(1) + c = a + b + c$. Since $2b = a+c$, $a+c = 2b$, so $a+b+c = 3b$ (not necessarily 0). Wait, let's check $x = 1$: $a + b + c = a + \frac{a+c}{2} + c = \frac{3(a+c)}{2} = 3b \neq 0$. Let's test $x = -1$: $a - b + c = a - \frac{a+c}{2} + c = \frac{a+c}{2} - b = b - b = 0$. Thus $x = -1$ is always a root!

MCQ 5: If the equations $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$ ($a \neq b$) have a common root, then $a + b$ equals:

A $1$

B $-1$

C $0$

D $2$

Derivation & Analysis:

Let common root be $\alpha$. Then $\alpha^2 + a\alpha + b = 0$ and $\alpha^2 + b\alpha + a = 0$. Subtracting yields $(a-b)\alpha + (b-a) = 0 \implies (a-b)(\alpha - 1) = 0$. Since $a \neq b$, $\alpha = 1$. Substituting $\alpha = 1$ in $1 + a + b = 0 \implies a + b = -1$.

MCQ 6: The set of all real values of $x$ satisfying $x^2 - 5x + 6 < 0$ is:

A $(2, 3)$

B $(-\infty, 2) \cup (3, \infty)$

C $[2, 3]$

D $\mathbb{R}$

Derivation & Analysis:

Factorize: $(x-2)(x-3) < 0$. For a parabola opening upwards, the expression is negative between its roots. Hence $x \in (2, 3)$.

MCQ 7: If $p$ and $q$ are roots of $x^2 + px + q = 0$, where $q \neq 0$, then the values of $p$ and $q$ are respectively:

A $-1, 2$

B $1, -2$

C $-2, 1$

D $2, -1$

Derivation & Analysis:

Sum of roots $p + q = -p \implies 2p + q = 0$. Product of roots $pq = q \implies q(p - 1) = 0$. Since $q \neq 0$, $p = 1$. Then $2(1) + q = 0 \implies q = -2$. Thus $p = 1, q = -2$.

MCQ 8: The number of real solutions of the equation $x^2 - 3|x| + 2 = 0$ is:

A $2$

B $4$

C $1$

D $3$

Derivation & Analysis:

Substitute $y = |x| \ge 0$, giving $y^2 - 3y + 2 = 0 \implies (y-1)(y-2) = 0 \implies y = 1$ or $y = 2$. Thus $|x| = 1 \implies x = \pm 1$ and $|x| = 2 \implies x = \pm 2$. Total 4 real solutions.

MCQ 9: If the quadratic equation $ax^2 + bx + c = 0$ has imaginary roots, then for all real $x$, the expression $ax^2 + bx + c$ has:

A Same sign as $a$

B Opposite sign to $a$

C Always positive

D Always negative

Derivation & Analysis:

If roots are imaginary, $\Delta < 0$, meaning the parabola never intersects the $x$-axis. Thus, it lies entirely above or entirely below the $x$-axis, retaining the exact sign of leading coefficient $a$ for all $x \in \mathbb{R}$.

Q10

MCQ 10: If $\alpha, \beta$ are the roots of $x^2 - 2x + 4 = 0$, then the value of $\alpha^5 + \beta^5$ is:

A $32$

B $-32$

C $64$

D $-64$

Derivation & Analysis:

The roots of $x^2 - 2x + 4 = 0$ can be written in polar form: $x = 1 \pm i\sqrt{3} = 2 e^{\pm i\pi/3}$. Then $\alpha = 2 cis(\pi/3)$ and $\beta = 2 cis(-\pi/3)$. $\alpha^5 + \beta^5 = 32 [cis(5\pi/3) + cis(-5\pi/3)] = 32 [2 \cos(5\pi/3)] = 64 (1/2) = 32$? Wait, $\cos(5\pi/3) = 1/2$, so $64(1/2) = 32$. Let's re-verify via recurrence: $S_1 = 2, S_2 = 4, S_3 = 2(4) - 4(2) = 0, S_4 = 2(0) - 4(4) = -16, S_5 = 2(-16) - 4(0) = -32$. Correct option is (B) $-32$.

Common questions

Candidates often memorize the cross-multiplication determinant formula without first checking if the two quadratic equations are proportional. If $a_1/a_2 = b_1/b_2 = c_1/c_2$, both roots are common, rendering the standard single-root determinant identity inapplicable.

Neglecting the sign of the leading coefficient $a$ when opening intervals or determining ranges. Always ensure $a > 0$ before applying standard interior/exterior interval rules for quadratic inequalities.

Master standard transformations like replacing $x$ with $x-k$ or $1/x$, and use recurrence relations $S_n = -b S_{n-1} - c S_{n-2}$ to evaluate higher-power sum of roots without tedious binomial expansions.

← Previous Topic

Theory of equations of degree ≥ 2

Next Topic →

Symmetric functions of roots