Quadratic Equations

Quadratic Equations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Standard Form: $ax^2 + bx + c = 0$ where $a \neq 0$.

Quadratic Formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Discriminant $D = b^2 - 4ac$:

• $D > 0$: Two distinct real roots
• $D = 0$: Two equal real roots
• $D < 0$: Complex conjugate roots (no real solutions)

Sum and Product of Roots: If roots are $\alpha$ and $\beta$: $$\alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a}$$

Nature of Roots:

• Both positive: $S > 0$ and $P > 0$ and $D \geq 0$
• Both negative: $S < 0$ and $P > 0$ and $D \geq 0$
• Opposite signs: $P < 0$ and $D \geq 0$
• Equal magnitude opposite sign: $S = 0$

⚡ JEE Tip: For questions about roots being in arithmetic progression (AP), geometric progression (GP), or harmonic progression (HP), form the equation using sum and product of roots. For AP: $2\beta = \alpha + \gamma$.

⚡ Common Mistake: When $a < 0$, the parabola opens downward. Don’t forget to check sign when finding maximum/minimum value.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Common Transformations:

If $\alpha, \beta$ are roots of $ax^2 + bx + c = 0$:

Equation	Roots	Sum	Product
$ax^2 + bx + c = 0$	$\alpha, \beta$	$-\frac{b}{a}$	$\frac{c}{a}$
$ax^2 - b’x + c = 0$ (with $b’ = \alpha+\beta$)	—	same as new	$\frac{c}{a}$
$cx^2 + bx + a = 0$	reciprocal $1/\alpha, 1/\beta$	$\frac{\beta+\alpha}{\alpha\beta}$	$\frac{1}{\alpha\beta}$
$a^2x^2 + b^2x + c^2 = 0$	squares $\alpha^2, \beta^2$	$\frac{b^2-2ac}{a^2}$	$\frac{c^2}{a^2}$

Maximum and Minimum Values:

For $y = ax^2 + bx + c$:

• Vertex occurs at $x = -\frac{b}{2a}$
• Value at vertex: $y_{\text{extreme}} = -\frac{D}{4a}$
• Maximum if $a < 0$, minimum if $a > 0$

Common Roots Problems:

Condition for common root between $f(x) = 0$ and $g(x) = 0$: Let the common root be $\alpha$. Then: $$\alpha = \frac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1} = \frac{c_1 b_2 - c_2 b_1}{b_1 a_2 - b_2 a_1}$$

Equating the two expressions for $\alpha$ gives the elimination (cross-multiplication) condition, which can also be obtained directly via the determinant method.

Worked Examples:

Example 1: The roots of $x^3 - 7x^2 + 14x - 8 = 0$ are in GP. Find them.

For three roots in GP, write them as $\dfrac{a}{r}, a, ar$. Their product equals $-\dfrac{\text{constant}}{\text{leading}} = 8$: $$\frac{a}{r}\cdot a \cdot ar = a^3 = 8 \implies a = 2.$$ So $x = 2$ is a root (the middle term). Dividing out $(x-2)$: $$x^3 - 7x^2 + 14x - 8 = (x-2)(x^2 - 5x + 4) = (x-2)(x-1)(x-4).$$ The roots are $1, 2, 4$, which form a GP with common ratio $r = 2$. This illustrates the key technique: when terms are in GP, choosing the symmetric parametrisation $a/r, a, ar$ makes the product collapse to $a^3$, isolating the middle term immediately.

Example 2 (JEE 2022): Find $m$ such that the roots of $x^2 + (3m-1)x + 2m^2 = 0$ are real and positive.

Real roots require $D \geq 0$: $$(3m-1)^2 - 8m^2 \geq 0 \implies 9m^2 - 6m + 1 - 8m^2 \geq 0 \implies m^2 - 6m + 1 \geq 0.$$ This gives $m \leq 3 - 2\sqrt{2}$ or $m \geq 3 + 2\sqrt{2}$.

Positive roots require $S > 0$ and $P > 0$:

• $S = -(3m-1) > 0 \implies m < \tfrac{1}{3}$.
• $P = 2m^2 > 0$, which holds for all $m \neq 0$.

Combining $m < \tfrac{1}{3}$ with the real-root condition (and noting $3 - 2\sqrt{2} \approx 0.172 < \tfrac{1}{3}$) leaves $m \leq 3 - 2\sqrt{2}$. At $m = 0$ the equation becomes $x^2 - x = 0$ with roots $0$ and $1$; since $0$ is not positive, exclude it.

Answer: $m \leq 3 - 2\sqrt{2}$ with $m \neq 0$.

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for serious JEE Advanced preparation.

Descartes’ Rule of Signs: The number of positive real roots equals the number of sign changes in $f(x)$, minus an even number. The number of negative real roots equals the number of sign changes in $f(-x)$, minus an even number.

Sturm’s Theorem: Provides the exact count of distinct real roots of a polynomial in a given interval via the Sturm sequence.

Rolle’s Theorem Applications: Between any two roots of $f(x) = 0$ there is at least one root of $f’(x) = 0$. If $f(x) = ax^2 + bx + c = 0$ has roots $\alpha < \beta$, then $f’(x) = 2ax + b = 0$ has its root at $$\gamma = -\frac{b}{2a} = \frac{\alpha + \beta}{2},$$ exactly the midpoint of the two roots.

Advanced Problems:

Problem: Both roots of $x^2 - px + q = 0$ are positive integers. Find the number of ordered pairs $(p, q)$ with $p, q \in [1, 10]$.

For $x^2 - px + q = 0$, the sum of roots is $p$ and the product is $q$. If the roots are positive integers $m, n$, then $m + n = p$ and $mn = q$. (Note: with $x^2 + px + q = 0$ the sum would be $-p < 0$, impossible for positive integer roots, so the minus sign is required.)

Enumerate unordered pairs $1 \leq m \leq n$ with product $q = mn \leq 10$ and sum $p = m + n \leq 10$:

$(m,n)$	$(p, q)$	Valid?
$(1,1)$	$(2,1)$	✓
$(1,2)$	$(3,2)$	✓
$(1,3)$	$(4,3)$	✓
$(1,4)$	$(5,4)$	✓
$(1,5)$	$(6,5)$	✓
$(1,6)$	$(7,6)$	✓
$(1,7)$	$(8,7)$	✓
$(1,8)$	$(9,8)$	✓
$(1,9)$	$(10,9)$	✓
$(1,10)$	$(11,10)$	reject ($p>10$)
$(2,2)$	$(4,4)$	✓
$(2,3)$	$(5,6)$	✓
$(2,4)$	$(6,8)$	✓
$(2,5)$	$(7,10)$	✓
$(3,3)$	$(6,9)$	✓
$(3,4)$	$(7,12)$	reject ($q>10$)

Each valid $(m,n)$ yields a distinct $(p,q)$, giving 14 ordered pairs $(p,q)$.

JEE Advanced Patterns (2018–2024):

• Transformation of roots (reciprocals, squares) is a recurring theme.
• Conditions for one root to be greater than or less than a given number.
• Common-root problems appear in mixed-difficulty sets.
• Location of roots (between intervals, in specific quadrants) tested in 2020 and 2023.
• Maxima–minima with quadratic constraints is trending upward.

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