Quadratic Equations - Solving quadratic equations by factorization, quadratic formula, and nature of roots

Hello young mathematicians! Welcome to this step-by-step tutorial on Chapter 4: Quadratic Equations from your Class 10 NCERT Mathematics syllabus.

Have you ever thrown a basketball into a hoop? The curved path that the ball follows through the air is called a parabola, and its height at any point can be calculated using a quadratic equation! Whether you are calculating the area of a rectangular garden, determining the speed of a stream, or designing bridges, quadratic equations are working behind the scenes.

In this guide, we will break down this chapter into simple, bite-sized concepts so that you can tackle any board exam question with confidence!

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Section 1: What is a Quadratic Equation?

In simple terms, a quadratic equation is a polynomial equation of degree 2. That means the highest exponent (power) of the variable is $2$.

The Standard Form

The standard form of a quadratic equation in variable $x$ is:

$$ax^2 + bx + c = 0$$

Where:

$x$ is the unknown variable.

$a, b,$ and $c$ are real numbers.

Crucial Condition: $a \neq 0$ (If $a = 0$, the $x^2$ term disappears, and it becomes a linear equation!).

Quick Check Examples:

$2x^2 - 5x + 3 = 0$ $\rightarrow$ Yes! Standard quadratic equation ($a=2, b=-5, c=3$).

$x^2 - 9 = 0$ $\rightarrow$ Yes! Here, $b=0$, which is completely allowed ($a=1, b=0, c=-9$).

$x + \frac{1}{x} = 2$ $\rightarrow$ Yes! If you multiply the entire equation by $x$, it becomes $x^2 + 1 = 2x \implies x^2 - 2x + 1 = 0$.

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Section 2: Method 1 – Solving by Factorization (Splitting the Middle Term)

Solving a quadratic equation means finding the values of $x$ (called the roots or zeros) that satisfy the equation. Since the degree is 2, a quadratic equation will always have at most two roots.

The Factorization Method works by splitting the linear middle term ($bx$) into two terms such that we can rewrite the equation as a product of two linear factors.

The 4-Step Recipe:

Write the equation in standard form: $ax^2 + bx + c = 0$.

Calculate the product $P = a \times c$ and note the sum $S = b$.

Find two numbers, $p$ and $q$, such that:

$p + q = b$

$p \times q = a \times c$

Rewrite $bx$ as $px + qx$, group the terms into two pairs, take common factors out, and set each linear factor to zero.

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Worked Example 1: Factorization Method

Solve the equation: $x^2 - 5x + 6 = 0$

Step 1: Compare with $ax^2 + bx + c = 0$.

Here, $a = 1$, $b = -5$, $c = 6$.

Step 2: Find product and sum.

Product ($a \times c$) = $1 \times 6 = 6$

Sum ($b$) = $-5$

Step 3: Find two numbers that multiply to $+6$ and add up to $-5$.

The numbers are $-2$ and $-3$ (because $(-2) \times (-3) = 6$ and $(-2) + (-3) = -5$).

Step 4: Split the middle term and factorize:

$$x^2 - 2x - 3x + 6 = 0$$

Group terms in pairs:

$$x(x - 2) - 3(x - 2) = 0$$

Take $(x - 2)$ common:

$$(x - 2)(x - 3) = 0$$

Step 5: Equate each factor to zero:

$x - 2 = 0 \implies x = 2$

$x - 3 = 0 \implies x = 3$

Answer: The roots of the equation are $x = 2$ and $x = 3$.

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Section 3: Method 2 – Solving using the Quadratic Formula

Sometimes, splitting the middle term can be tricky, especially when square roots or large numbers are involved. Don't worry! Ancient Indian mathematician Sridharacharya gave us a foolproof formula that directly yields the roots.

The Quadratic Formula:

For any quadratic equation $ax^2 + bx + c = 0$ (where $a \neq 0$):

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

The symbol $\pm$ (plus-minus) means you will calculate two values: one using $+$ and one using $-$.

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Worked Example 2: Quadratic Formula

Solve the equation: $2x^2 - 7x + 3 = 0$

Step 1: Identify $a, b,$ and $c$.

$a = 2$, $b = -7$, $c = 3$

Step 2: Substitute into the formula.

$$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)}$$

Step 3: Simplify step-by-step.

$$x = \frac{7 \pm \sqrt{49 - 24}}{4}$$

$$x = \frac{7 \pm \sqrt{25}}{4}$$

$$x = \frac{7 \pm 5}{4}$$

Step 4: Separate the two roots.

Root 1 ($+$ case): $x = \frac{7 + 5}{4} = \frac{12}{4} = 3$

Root 2 ($-$ case): $x = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}$

Answer: The roots are $x = 3$ and $x = \frac{1}{2}$.

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Section 4: Nature of Roots (The Magic of Discriminant $D$)

Have you noticed the expression inside the square root in the formula? That expression, $b^2 - 4ac$, is called the Discriminant (denoted by $D$).

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

The Discriminant determines the nature of the roots without requiring you to fully solve the equation!

Value of Discriminant ($D$)	Nature of Roots	What the roots look like
$D > 0$ (Positive)	Two distinct (different) real roots	$x = \frac{-b + \sqrt{D}}{2a}$ and $x = \frac{-b - \sqrt{D}}{2a}$
$D = 0$ (Zero)	Two equal real roots	$x = -\frac{b}{2a}$ (both roots are the same)
$D < 0$ (Negative)	No real roots (Imaginary roots)	Square root of a negative number is not real

💡 Teacher's Tip for Exams: If a board question asks you to "Determine whether real roots exist," always calculate $D = b^2 - 4ac$ first!

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Practice Time! (3 Board-Style Questions with Solutions)

Now it's your turn to test your understanding! Try solving these on paper before reading the step-by-step solutions below.

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Question 1 (Factorization Method)

Solve for $x$: $x^2 - 3x - 10 = 0$

<summary>🔍 Click to view Solution</summary>

Solution:

Given equation: $x^2 - 3x - 10 = 0$

$a = 1$, $b = -3$, $c = -10$

Product = $1 \times (-10) = -10$

Sum = $-3$

We need two numbers whose product is $-10$ and sum is $-3$.

The numbers are $-5$ and $+2$.

Splitting the middle term:

$$x^2 - 5x + 2x - 10 = 0$$

Group terms:

$$x(x - 5) + 2(x - 5) = 0$$

$$(x - 5)(x + 2) = 0$$

Equating factors to zero:

$x - 5 = 0 \implies x = 5$

$x + 2 = 0 \implies x = -2$

Final Answer: $x = 5$ and $x = -2$

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Question 2 (Quadratic Formula)

Solve for $x$ using the Quadratic Formula: $2x^2 + x - 528 = 0$

<summary>🔍 Click to view Solution</summary>

Solution:

Given equation: $2x^2 + x - 528 = 0$

Here $a = 2$, $b = 1$, $c = -528$

Step 1: Calculate the Discriminant ($D$)

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

$$D = (1)^2 - 4(2)(-528)$$

$$D = 1 + 4224 = 4225$$

Since $D > 0$, two distinct real roots exist!

Note that $\sqrt{4225} = 65$.

Step 2: Apply the Quadratic Formula

$$x = \frac{-b \pm \sqrt{D}}{2a}$$

$$x = \frac{-1 \pm 65}{2(2)} = \frac{-1 \pm 65}{4}$$

Step 3: Calculate the roots

$x = \frac{-1 + 65}{4} = \frac{64}{4} = 16$

$x = \frac{-1 - 65}{4} = \frac{-66}{4} = -\frac{33}{2}$

Final Answer: $x = 16$ and $x = -\frac{33}{2}$

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Question 3 (Nature of Roots / Finding Unknown $k$)

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

<summary>🔍 Click to view Solution</summary>

Solution:

Given equation: $2x^2 + kx + 3 = 0$

Here $a = 2$, $b = k$, $c = 3$

Condition for Equal Roots:

A quadratic equation has two equal real roots if and only if its Discriminant is zero ($D = 0$).

$$D = b^2 - 4ac = 0$$

Substitute the values:

$$k^2 - 4(2)(3) = 0$$

$$k^2 - 24 = 0$$

$$k^2 = 24$$

$$k = \pm \sqrt{24}$$

$$k = \pm 2\sqrt{6}$$

Final Answer: The required value of $k$ is $\pm 2\sqrt{6}$ (i.e., $2\sqrt{6}$ or $-2\sqrt{6}$).

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Quick Revision Summary

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

Methods to Solve:

Factorization: Split $bx$ into two numbers that add up to $b$ and multiply to $ac$.

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

Nature of Roots ($D = b^2 - 4ac$):

$D > 0 \implies$ Two real and distinct roots.

$D = 0 \implies$ Two real and equal roots.

$D < 0 \implies$ No real roots.

Keep practicing regularly, and remember: math isn't about memorization—it's about practice and understanding the logic. You've got this! Happy learning!