JEE Maths — Quadratic Equations

Chapter Overview & Weightage

Quadratic Equations is a foundational JEE topic that links to Sequences, Complex Numbers, and Coordinate Geometry. JEE Main typically asks 1-2 direct questions; JEE Advanced uses quadratics inside multi-concept problems. Investing in this chapter pays off across the entire syllabus.

Year	JEE Main Qs	JEE Adv Qs
2024	2	1
2023	1	2
2022	2	1
2021	2	1
2020	1	2

Average JEE Main weightage: 8 marks per session. Average JEE Advanced: 4-6 marks. Worth at least one solid week of focused practice in your final months.

Key Concepts You Must Know

Standard form: $ax^2 + bx + c = 0$ , $a \ne 0$ .
Roots formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
Discriminant $D = b^2 - 4ac$ : $D > 0$ real distinct, $D = 0$ real equal, $D < 0$ complex conjugate.
Sum and product of roots: $\alpha + \beta = -b/a$ , $\alpha\beta = c/a$ .
Nature of roots based on coefficients.
Common roots: condition for two quadratics to share a root.
Roots in given intervals: location of roots using $f(x_0)$ , vertex position.
Symmetric functions of roots: $\alpha^2 + \beta^2$ , $\alpha^3 + \beta^3$ , etc.

Important Formulas

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

$\alpha - \beta = \pm\frac{\sqrt{D}}{a}, \quad D = b^2 - 4ac$

These are JEE-Main bread and butter — coefficients to root-symmetric-functions in one step.

Two quadratics $a_1 x^2 + b_1 x + c_1 = 0$ and $a_2 x^2 + b_2 x + c_2 = 0$ share both roots iff $a_1/a_2 = b_1/b_2 = c_1/c_2$ .

Share exactly one root iff $(a_1 c_2 - a_2 c_1)^2 = (a_1 b_2 - a_2 b_1)(b_1 c_2 - b_2 c_1)$ .

For both roots of $f(x) = ax^2 + bx + c$ to lie in $(x_1, x_2)$ (assuming $a > 0$ ):

$D \geq 0$
$a f(x_1) > 0$ and $a f(x_2) > 0$
$x_1 < -b/(2a) < x_2$

Solved Previous Year Questions

PYQ 1 (JEE Main 2024)

If the roots of $x^2 - 5x + k = 0$ are in the ratio $2:3$ , find $k$ .

Let the roots be $2\alpha$ and $3\alpha$ . Sum: $5\alpha = 5$ , so $\alpha = 1$ . Roots are $2$ and $3$ .

Product: $k = 2 \times 3 = 6$ .

PYQ 2 (JEE Advanced 2023)

Find the values of $a$ for which $(a^2 - 1)x^2 + 2(a - 1)x + 2 = 0$ has both roots positive.

For both roots positive: (i) $D \geq 0$ , (ii) sum $> 0$ , (iii) product $> 0$ .

$D = 4(a-1)^2 - 8(a^2 - 1) = 4(a-1)[(a-1) - 2(a+1)] = 4(a-1)(-a - 3) \geq 0$ .

So $(a - 1)(a + 3) \leq 0$ , giving $-3 \leq a \leq 1$ .

Sum $-2(a-1)/(a^2-1) = -2/(a+1) > 0 \implies a < -1$ .

Product $2/(a^2-1) > 0 \implies a^2 > 1 \implies a < -1$ or $a > 1$ .

Combining: $-3 \leq a < -1$ .

PYQ 3 (JEE Main 2022)

If $\alpha, \beta$ are roots of $x^2 + 7x + 12 = 0$ , find $\alpha^3 + \beta^3$ .

$\alpha + \beta = -7$ , $\alpha\beta = 12$ . Use:

$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = -343 - 3(12)(-7) = -343 + 252 = -91$

Difficulty Distribution

Sub-topic	Easy	Medium	Hard
Direct roots	60%	35%	5%
Symmetric functions	30%	50%	20%
Common roots	20%	50%	30%
Interval problems	10%	40%	50%

Interval problems separate Advanced-level students from Main-level ones.

Expert Strategy

For “find $k$ such that…” questions, always state your three conditions ( $D$ , sign of $f(x_1)$ , vertex position) explicitly. JEE Advanced often uses partial-credit grading on these.

When dealing with $\alpha^n + \beta^n$ , use the recurrence $S_n = (\alpha + \beta) S_{n-1} - (\alpha\beta) S_{n-2}$ . Faster than expanding by hand for $n \geq 4$ .

If the coefficient of $x^2$ contains a parameter ( $a^2 - 1$ above), check whether the equation is even quadratic — at $a = \pm 1$ it becomes linear. JEE Advanced traps you here every year.

Common Traps

Forgetting that $a$ can be zero (or contain a parameter that goes to zero), turning the equation linear. Always check the leading coefficient.

Using only $D \geq 0$ for “real roots” without checking which side they fall on. The “both roots positive” or “both roots in $(0, 1)$ ” conditions need the full set of three inequalities.

Confusing “roots equal” ( $D = 0$ , repeated root) with “roots common” (two equations share a root). They are different setups.