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Quadratic Equations formulas For JEE 2026
Quadratic Equations is an important chapter in JEE Mathematics and is asked regularly in the exam, usually with 1–2 questions every year. This topic includes key concepts like the quadratic formula, discriminant, and the nature of roots, which help students solve different types of equations easily. It also covers Vieta’s formulas and symmetric functions of roots, which are useful for understanding the relationship between the roots.
Students should also practice quadratic inequalities and conditions on roots, as these are commonly tested in JEE. With a clear understanding of formulas and regular practice, this chapter can become a scoring area. For quick revision, students can also refer to a well-organized JEE Mains Maths Formula PDF to revise important formulas and concepts effectively.
What Is a Quadratic Equation?
A quadratic equation is a polynomial equation of degree 2 — the highest power of the unknown variable is 2. These equations arise everywhere in mathematics and physics (projectile motion, area problems, optimization), and solving them is a core JEE skill.
Quadratic Equation: An equation of the form $$ax^2 + bx + c = 0$$, where $$a$$, $$b$$, $$c$$ are real numbers and $$a \neq 0$$. Here $$a$$ is the leading coefficient, $$b$$ is the coefficient of $$x$$, and $$c$$ is the constant term.
Note: The condition $$a \neq 0$$ is essential. If $$a = 0$$, the equation becomes linear ($$bx + c = 0$$), not quadratic.
Discriminant and Nature of Roots Formulas
Before solving a quadratic equation, we can determine the nature of its roots without actually finding them. The key is a quantity called the discriminant.
Discriminant: For the quadratic $$ax^2 + bx + c = 0$$, the discriminant is $$D = b^2 - 4ac$$.
Nature of Roots Based on Discriminant
For $$ax^2 + bx + c = 0$$ with $$a, b, c \in \mathbb{R}$$ and $$a \neq 0$$:
| Discriminant | Nature of Roots |
|---|---|
| $$D > 0$$ | Two distinct real roots |
| $$D = 0$$ | Two equal real roots (repeated root) |
| $$D < 0$$ | Two complex conjugate roots (no real roots) |
Further refinements (when $$a, b, c \in \mathbb{Q}$$):
- $$D > 0$$ and $$D$$ is a perfect square $$\Rightarrow$$ roots are rational
- $$D > 0$$ and $$D$$ is not a perfect square $$\Rightarrow$$ roots are irrational (and occur in conjugate pairs $$p + \sqrt{q}$$, $$p - \sqrt{q}$$)
Worked Example
Determine the nature of roots of $$2x^2 - 5x + 3 = 0$$.
$$a = 2$$, $$b = -5$$, $$c = 3$$
$$D = b^2 - 4ac = (-5)^2 - 4(2)(3) = 25 - 24 = 1$$
Since $$D = 1 > 0$$ and $$1$$ is a perfect square, the equation has two distinct rational roots.
Quadratic Formula
The quadratic formula gives the exact roots of any quadratic equation directly from its coefficients. It is derived by completing the square on the general form $$ax^2 + bx + c = 0$$.
The roots of $$ax^2 + bx + c = 0$$ are:
$$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{D}}{2a}$$$
The two roots are often written as:
$$$\alpha = \frac{-b + \sqrt{D}}{2a}, \quad \beta = \frac{-b - \sqrt{D}}{2a}$$$
Worked Example
Solve $$x^2 - 7x + 10 = 0$$.
$$a = 1$$, $$b = -7$$, $$c = 10$$
$$D = (-7)^2 - 4(1)(10) = 49 - 40 = 9$$
$$x = \frac{-(-7) \pm \sqrt{9}}{2(1)} = \frac{7 \pm 3}{2}$$
$$x = \frac{7 + 3}{2} = 5$$ or $$x = \frac{7 - 3}{2} = 2$$
Roots: $$x = 5$$ and $$x = 2$$.
Worked Example
Solve $$2x^2 + 3x + 5 = 0$$.
$$D = 3^2 - 4(2)(5) = 9 - 40 = -31 < 0$$
$$x = \frac{-3 \pm \sqrt{-31}}{4} = \frac{-3 \pm i\sqrt{31}}{4}$$
The roots are complex: $$x = \dfrac{-3 + i\sqrt{31}}{4}$$ and $$x = \dfrac{-3 - i\sqrt{31}}{4}$$.
Note that they form a conjugate pair.
Tip: If the coefficients are integers and you can spot factors, factorization is faster than the formula. For example, $$x^2 - 7x + 10 = (x - 5)(x - 2) = 0$$ gives roots 5 and 2 immediately.
Vieta's Formulas (Sum and Product of Roots)
Even without solving a quadratic, we can find the sum and product of its roots directly from the coefficients. These relations, known as Vieta's formulas, are among the most powerful and frequently tested tools in JEE.
Vieta's Formulas for Quadratics
If $$\alpha$$ and $$\beta$$ are the roots of $$ax^2 + bx + c = 0$$, then:
$$$\alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a}$$$
Memory aid: Sum of roots $$= \dfrac{-(\text{coefficient of } x)}{\text{coefficient of } x^2}$$, Product of roots $$= \dfrac{\text{constant term}}{\text{coefficient of } x^2}$$
Worked Example
If $$\alpha, \beta$$ are roots of $$3x^2 - 12x + 7 = 0$$, find $$\alpha + \beta$$ and $$\alpha\beta$$.
$$\alpha + \beta = -\frac{-12}{3} = \frac{12}{3} =$$ 4
$$\alpha\beta = \frac{7}{3} =$$ $$\dfrac{7}{3}$$
Symmetric Functions of Roots
Many JEE problems ask you to find expressions involving $$\alpha$$ and $$\beta$$ (like $$\alpha^2 + \beta^2$$ or $$\frac{1}{\alpha} + \frac{1}{\beta}$$) without finding the roots themselves. These can always be expressed using $$\alpha + \beta$$ and $$\alpha\beta$$.
Useful Identities for Symmetric Functions
- $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$
- $$\alpha^2 - \beta^2 = (\alpha + \beta)(\alpha - \beta)$$
- $$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = \dfrac{D}{a^2}$$
- $$|\alpha - \beta| = \dfrac{\sqrt{D}}{|a|}$$
- $$\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta}$$
- $$\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{\alpha^2 + \beta^2}{\alpha\beta}$$
- $$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$$
- $$\alpha^3 - \beta^3 = (\alpha - \beta)(\alpha^2 + \alpha\beta + \beta^2)$$
Worked Example
If $$\alpha, \beta$$ are roots of $$x^2 - 5x + 3 = 0$$, find $$\alpha^2 + \beta^2$$ and $$\dfrac{1}{\alpha} + \dfrac{1}{\beta}$$.
$$\alpha + \beta = 5$$, $$\alpha\beta = 3$$
$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 25 - 6 =$$ 19
$$\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{5}{3} =$$ $$\dfrac{5}{3}$$
Forming Quadratic Equations from Roots
If we know the roots of a quadratic, we can reconstruct the equation. This is the reverse of solving — instead of going from equation to roots, we go from roots to equation.
If $$\alpha$$ and $$\beta$$ are the roots, then the quadratic equation is:
$$$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$$
or equivalently: $$(x - \alpha)(x - \beta) = 0$$
Worked Example
Form a quadratic equation whose roots are 3 and $$-2$$.
Sum of roots $$= 3 + (-2) = 1$$
Product of roots $$= 3 \times (-2) = -6$$
Equation: $$x^2 - (1)x + (-6) = 0$$
$$x^2 - x - 6 = 0$$
Verify: $$(x - 3)(x + 2) = x^2 - x - 6$$ ✓
Worked Example
If $$\alpha, \beta$$ are roots of $$x^2 - 3x + 1 = 0$$, form the equation whose roots are $$\alpha^2$$ and $$\beta^2$$.
From the original: $$\alpha + \beta = 3$$, $$\alpha\beta = 1$$.
Sum of new roots: $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 9 - 2 = 7$$
Product of new roots: $$\alpha^2\beta^2 = (\alpha\beta)^2 = 1$$
New equation: $$x^2 - 7x + 1 = 0$$
Common Roots of Two Quadratic Equations
Condition for Common Roots
For $$a_1x^2 + b_1x + c_1 = 0$$ and $$a_2x^2 + b_2x + c_2 = 0$$:
One common root ($$\alpha$$):
- The common root satisfies: $$\alpha = \dfrac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}$$
- Condition: $$(c_1a_2 - c_2a_1)^2 = (a_1b_2 - a_2b_1)(b_1c_2 - b_2c_1)$$
Both roots common:
$$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$$
Worked Example
Find the value of $$k$$ if $$x^2 + 2x + 3 = 0$$ and $$x^2 + kx + 5 = 0$$ have a common root.
Let $$\alpha$$ be the common root. Then:
$$\alpha^2 + 2\alpha + 3 = 0$$ … (1)
$$\alpha^2 + k\alpha + 5 = 0$$ … (2)
Subtracting (1) from (2): $$(k - 2)\alpha + 2 = 0$$, so $$\alpha = \frac{-2}{k - 2}$$
Substituting back into (1) and simplifying gives $$4t^2 - 4t + 3 = 0$$ (where $$t = \frac{1}{k-2}$$).
$$D = 16 - 48 = -32 < 0$$
Since the discriminant is negative, there is no real value of $$k$$ for which the equations have a common root.
Maximum and Minimum of Quadratic Expressions
The quadratic expression $$f(x) = ax^2 + bx + c$$ represents a parabola. If $$a > 0$$, the parabola opens upward and has a minimum. If $$a < 0$$, it opens downward and has a maximum.
Vertex: The turning point of the parabola $$y = ax^2 + bx + c$$, located at $$x = -\dfrac{b}{2a}$$.
Extreme Values of $$ax^2 + bx + c$$
The extreme value occurs at $$x = -\dfrac{b}{2a}$$, and the extreme value is $$-\dfrac{D}{4a}$$ where $$D = b^2 - 4ac$$.
| Condition | Parabola shape | Extreme value |
|---|---|---|
| $$a > 0$$ | Opens upward | Minimum value $$= -\dfrac{D}{4a}$$ |
| $$a < 0$$ | Opens downward | Maximum value $$= -\dfrac{D}{4a}$$ |
- If $$a > 0$$: $$f(x) \geq -\dfrac{D}{4a}$$ for all $$x$$ (range: $$\left[-\dfrac{D}{4a}, \infty\right)$$)
- If $$a < 0$$: $$f(x) \leq -\dfrac{D}{4a}$$ for all $$x$$ (range: $$\left(-\infty, -\dfrac{D}{4a}\right]$$)
Worked Example
Find the minimum value of $$f(x) = 2x^2 - 8x + 5$$.
$$a = 2 > 0$$, so the parabola opens upward and has a minimum.
$$D = (-8)^2 - 4(2)(5) = 64 - 40 = 24$$
Minimum value $$= -\dfrac{D}{4a} = -\dfrac{24}{8} =$$ $$-3$$
This occurs at $$x = -\dfrac{b}{2a} = -\dfrac{-8}{4} = 2$$.
Verify: $$f(2) = 2(4) - 8(2) + 5 = 8 - 16 + 5 = -3$$ ✓
Sign of a Quadratic Expression
Knowing when a quadratic expression is positive, negative, or zero is essential for solving inequalities.
Sign Analysis of $$f(x) = ax^2 + bx + c$$
Let $$\alpha \leq \beta$$ be the roots (when they are real).
Case 1: $$D > 0$$ (two distinct real roots $$\alpha < \beta$$)
- If $$a > 0$$: $$f(x) > 0$$ for $$x < \alpha$$ or $$x > \beta$$; $$f(x) < 0$$ for $$\alpha < x < \beta$$
- If $$a < 0$$: $$f(x) < 0$$ for $$x < \alpha$$ or $$x > \beta$$; $$f(x) > 0$$ for $$\alpha < x < \beta$$
Case 2: $$D = 0$$ (repeated root $$\alpha$$)
- $$f(x) = a(x - \alpha)^2$$, so $$f(x)$$ has the same sign as $$a$$ for all $$x \neq \alpha$$, and $$f(\alpha) = 0$$
Case 3: $$D < 0$$ (no real roots)
- $$f(x)$$ has the same sign as $$a$$ for all real $$x$$ (the parabola never crosses the $$x$$-axis)
Note: If $$a > 0$$ and $$D < 0$$, then $$f(x) > 0$$ for all real $$x$$ (the expression is always positive). If $$a < 0$$ and $$D < 0$$, then $$f(x) < 0$$ for all real $$x$$.
Worked Example
For what values of $$k$$ is $$x^2 + kx + 4 > 0$$ for all real $$x$$?
For $$f(x) = x^2 + kx + 4$$ to be always positive:
Condition 1: $$a = 1 > 0$$ ✓
Condition 2: $$D < 0$$
$$D = k^2 - 4(1)(4) = k^2 - 16 < 0$$
$$k^2 < 16$$
$$-4 < k < 4$$
Quadratic Inequalities Formula
Solving Quadratic Inequalities
To solve $$ax^2 + bx + c > 0$$ (or $$< 0$$, $$\geq 0$$, $$\leq 0$$):
Step 1: Find the roots $$\alpha, \beta$$ (with $$\alpha \leq \beta$$) using the quadratic formula.
Step 2: Use the sign chart:
If $$a > 0$$:
- $$ax^2 + bx + c > 0 \implies x \in (-\infty, \alpha) \cup (\beta, \infty)$$
- $$ax^2 + bx + c < 0 \implies x \in (\alpha, \beta)$$
If $$a < 0$$:
- $$ax^2 + bx + c > 0 \implies x \in (\alpha, \beta)$$
- $$ax^2 + bx + c < 0 \implies x \in (-\infty, \alpha) \cup (\beta, \infty)$$
For $$\geq$$ or $$\leq$$, include the endpoints.
Worked Example
Solve $$x^2 - 5x + 6 \leq 0$$.
Step 1: Find roots. $$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$$, so $$\alpha = 2$$, $$\beta = 3$$.
Step 2: Since $$a = 1 > 0$$, the expression is $$\leq 0$$ between the roots.
Solution: $$x \in [2, 3]$$
Worked Example
Solve $$-2x^2 + 7x - 3 > 0$$.
Multiply by $$-1$$ (flip inequality): $$2x^2 - 7x + 3 < 0$$
Find roots: $$D = 49 - 24 = 25$$
$$x = \frac{7 \pm 5}{4}$$, so $$x = 3$$ or $$x = \frac{1}{2}$$
Since $$a = 2 > 0$$, the expression $$2x^2 - 7x + 3 < 0$$ when $$\frac{1}{2} < x < 3$$.
Solution: $$x \in \left(\dfrac{1}{2}, 3\right)$$
Tip: Quick approach for JEE: if $$a > 0$$, then $$f(x) < 0$$ "between the roots" and $$f(x) > 0$$ "outside the roots." Draw a rough upward parabola to visualize.
Conditions on Location of Roots
JEE problems often specify conditions on the roots (e.g., both roots positive, roots in a given interval, one root greater than $$k$$). These translate into conditions on the coefficients.
For $$f(x) = ax^2 + bx + c$$ with $$a > 0$$ and roots $$\alpha, \beta$$:
Both roots positive: $$D \geq 0$$, $$\alpha + \beta > 0$$, $$\alpha\beta > 0$$
Both roots negative: $$D \geq 0$$, $$\alpha + \beta < 0$$, $$\alpha\beta > 0$$
Roots of opposite signs: $$\alpha\beta < 0$$ (i.e., $$\frac{c}{a} < 0$$; discriminant is automatically positive)
Both roots greater than $$k$$: $$D \geq 0$$, $$f(k) > 0$$, $$-\frac{b}{2a} > k$$
Both roots less than $$k$$: $$D \geq 0$$, $$f(k) > 0$$, $$-\frac{b}{2a} < k$$
$$k$$ lies between the roots: $$f(k) < 0$$ (only this one condition is needed when $$a > 0$$)
Worked Example
Find the values of $$k$$ for which both roots of $$x^2 - 6x + k = 0$$ are positive.
$$a = 1$$, $$b = -6$$, $$c = k$$.
Condition 1: $$D \geq 0 \implies 36 - 4k \geq 0 \implies k \leq 9$$
Condition 2: $$\alpha + \beta > 0 \implies 6 > 0$$ ✓ (always true)
Condition 3: $$\alpha\beta > 0 \implies k > 0$$
Combining: $$0 < k \leq 9$$
Worked Example
Find the values of $$k$$ for which 2 lies between the roots of $$x^2 - (k+1)x + (k-1) = 0$$.
For $$a = 1 > 0$$ and 2 between the roots, we need $$f(2) < 0$$:
$$f(2) = 4 - 2(k+1) + (k-1) = 4 - 2k - 2 + k - 1 = 1 - k$$
$$1 - k < 0 \implies k > 1$$
Solution: $$k > 1$$
Tip: The condition "$$k$$ lies between the roots" for $$a > 0$$ requires only $$f(k) < 0$$. This is the simplest and most elegant condition — no need to check discriminant or vertex separately. JEE frequently tests this shortcut
Quadratic Equations formulas For JEE 2026: Conclusion
Quadratic Equations form a strong foundation in JEE Mathematics and are widely used in solving a variety of problems. Understanding concepts like discriminant, roots, and Vieta’s formulas helps in simplifying complex questions and improving accuracy.
Regular practice of inequalities, root-based conditions, and graphical interpretation strengthens problem-solving skills. With consistent revision and application of formulas, this chapter can become a reliable scoring area in the exam
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