Chapter 4

Class 10 Mathematics - Chapter 4: Quadratic Equations
Chapter 4: Quadratic Equations
1. Standard Form of a Quadratic Equation
Definition: A quadratic equation in the variable $x$ is an equation of the form: $$ax^2 + bx + c = 0$$ where $a, b, c$ are real numbers and $a \neq 0$.

Roots of an Equation: A real number $\alpha$ is called a root of the quadratic equation $ax^2 + bx + c = 0$ if: $$a\alpha^2 + b\alpha + c = 0$$
Example 1.1: Checking Quadratic Form

Check whether $(x - 2)^2 + 1 = 2x - 3$ is a quadratic equation.

Solution:
Expanding the left side: $$x^2 - 4x + 4 + 1 = 2x - 3$$ $$x^2 - 4x + 5 = 2x - 3$$ Move all terms to the left side: $$x^2 - 4x - 2x + 5 + 3 = 0$$ $$x^2 - 6x + 8 = 0$$ Since it is of the form $ax^2 + bx + c = 0$ with $a = 1 \neq 0$, it is a quadratic equation.

2. Methods for Solving Quadratic Equations
Method 1: Factorisation (Splitting the Middle Term)
To solve $ax^2 + bx + c = 0$, find two numbers whose sum is $b$ and product is $ac$. Express the quadratic expression as a product of two linear factors and set each factor equal to zero.
Example 2.1: Solution by Factorisation

Find the roots of the equation $2x^2 - 5x + 3 = 0$.

Solution:
We need two numbers whose product is $2 \times 3 = 6$ and sum is $-5$. These numbers are $-2$ and $-3$. $$2x^2 - 2x - 3x + 3 = 0$$ $$2x(x - 1) - 3(x - 1) = 0$$ $$(2x - 3)(x - 1) = 0$$ Setting each factor to $0$: $$2x - 3 = 0 \implies x = \frac{3}{2}$$ $$x - 1 = 0 \implies x = 1$$ Therefore, the roots are $x = \frac{3}{2}$ and $x = 1$.

Method 2: Quadratic Formula
For any quadratic equation $ax^2 + bx + c = 0$ (where $a \neq 0$), if $b^2 - 4ac \geq 0$, the real roots are given by: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Example 2.2: Solution using Quadratic Formula

Solve the quadratic equation $3x^2 - 5x + 2 = 0$.

Solution:
Comparing with $ax^2 + bx + c = 0$, we have:
$a = 3$, $b = -5$, $c = 2$.

Calculate the Discriminant ($D$): $$D = b^2 - 4ac$$ $$D = (-5)^2 - 4(3)(2)$$ $$D = 25 - 24 = 1$$ Since $D > 0$, real roots exist. Applying the quadratic formula: $$x = \frac{-(-5) \pm \sqrt{1}}{2(3)}$$ $$x = \frac{5 \pm 1}{6}$$ Breaking into individual values on separate lines: $$x = \frac{5 + 1}{6} = \frac{6}{6} = 1$$ $$x = \frac{5 - 1}{6} = \frac{4}{6} = \frac{2}{3}$$ The roots are $x = 1$ and $x = \frac{2}{3}$.

3. Nature of Roots
The expression $D = b^2 - 4ac$ is called the Discriminant. It determines the nature of roots for $ax^2 + bx + c = 0$:
Discriminant Value ($D$) Nature of Roots Roots Value
$D > 0$ Two distinct real roots $x = \frac{-b \pm \sqrt{D}}{2a}$
$D = 0$ Two equal real roots $x = -\frac{b}{2a}, -\frac{b}{2a}$
$D < 0$ No real roots None (Real domain)
Example 3.1: Finding Unknown $k$ for Equal Roots

Find the value of $k$ for which the quadratic equation $2x^2 + kx + 3 = 0$ has two equal real roots.

Solution:
Here $a = 2$, $b = k$, and $c = 3$.
For equal roots, Discriminant $D = 0$: $$D = b^2 - 4ac = 0$$ $$k^2 - 4(2)(3) = 0$$ $$k^2 - 24 = 0$$ $$k^2 = 24$$ $$k = \pm\sqrt{24}$$ $$k = \pm 2\sqrt{6}$$ Thus, the required values are $k = \pm 2\sqrt{6}$.