Solution of Quadratic Equations by Quadratic Formula for CBSE Class 10

While factorisation by splitting the middle term is an elegant technique, it relies heavily on finding integer or simple rational factors of the product $ac$. For quadratic equations with large coefficients, fractions, or irrational roots—such as $x^2 + 4x - 1 = 0$—splitting the middle term becomes nearly impossible by inspection.

To overcome this limitation, ancient Indian mathematician Sridharacharya derived a universal algebraic formula that solves any quadratic equation directly from its coefficients. In CBSE Class 10 Mathematics, the Quadratic Formula is an indispensable tool that guarantees a solution whenever real roots exist.

What You Will Learn

• Historical background and Sridharacharya's rule
• Derivation of the Quadratic Formula using the Method of Completing the Square
• Statement and structure of the Quadratic Formula
• The role of the Discriminant ($D = b^2 - 4ac$) as the gatekeeper of real roots
• Step-by-step protocol for applying the formula without arithmetic mistakes
• Solved CBSE board examination problems and practical exam strategies

1. Derivation by Completing the Square

Consider the general quadratic equation in standard form: $ax^2 + bx + c = 0, \quad \text{where } a \ne 0$

Step-by-Step Derivation:

1. Divide throughout by the leading coefficient $a$: $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$
2. Transpose the constant term to the right-hand side: $x^2 + \frac{b}{a}x = -\frac{c}{a}$
3. Complete the square on the left-hand side: Notice that $\left(x + \frac{b}{2a}\right)^2 = x^2 + 2(x)\left(\frac{b}{2a}\right) + \left(\frac{b}{2a}\right)^2 = x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}$. To create a perfect square, add $\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$ to both sides: $x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a}$
4. Write the LHS as a perfect square and simplify the RHS: $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$
5. Take the square root of both sides: $x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$
6. Solve explicitly for $x$: $x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$ $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

2. The Quadratic Formula

For any quadratic equation $ax^2 + bx + c = 0$ with $a \ne 0$, its roots are given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ provided that $b^2 - 4ac \ge 0$.

Here, the quantity under the radical sign, $D = b^2 - 4ac$, is called the discriminant.

The two distinct roots are: $\alpha = \frac{-b + \sqrt{D}}{2a} \quad \text{and} \quad \beta = \frac{-b - \sqrt{D}}{2a}$

Important: <u>If $b^2 - 4ac < 0$, the quantity under the square root is negative. Since the square root of a negative number is not a real number, the quadratic equation has NO real roots.</u>

3. Step-by-Step Procedure for Applying the Formula

To avoid algebraic and sign errors, follow this structured four-step procedure:

1. Step 1: Write the given equation strictly in standard form $ax^2 + bx + c = 0$ and list the exact values of $a, b,$ and $c$ (including their signs!).
2. Step 2: Calculate the discriminant separately: $D = b^2 - 4ac$
3. Step 3: Check the sign of $D$:
   • If $D < 0$, stop and write: "Since $D < 0$, the equation has no real roots."
   • If $D \ge 0$, proceed to calculate $\sqrt{D}$.
4. Step 4: Substitute $a, b,$ and $\sqrt{D}$ into $x = \frac{-b \pm \sqrt{D}}{2a}$ and simplify to find both roots.

4. Solved CBSE Board Examination Problems

Solved Example 1: Equation Not Easily Factorable

Problem: Solve the quadratic equation $x^2 + 4x - 5 = 0$ using the quadratic formula.

Solution:

1. Compare with $ax^2 + bx + c = 0$: $a = 1, \quad b = 4, \quad c = -5$
2. Calculate the discriminant $D$: $D = b^2 - 4ac = (4)^2 - 4(1)(-5) = 16 + 20 = 36$
3. Since $D = 36 > 0$, real roots exist: $\sqrt{D} = \sqrt{36} = 6$
4. Apply the quadratic formula: $x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm 6}{2(1)}$
   • First root: $x = \frac{-4 + 6}{2} = \frac{2}{2} = 1$
   • Second root: $x = \frac{-4 - 6}{2} = \frac{-10}{2} = -5$
5. Therefore, <u>the roots are $x = 1$ and $x = -5$</u>.

Solved Example 2: Irrational Roots (Board Exam Favorite)

Problem: Solve for $x$: $2x^2 - 2\sqrt{2}x + 1 = 0$.

Solution:

1. Identify coefficients: $a = 2, b = -2\sqrt{2}, c = 1$.
2. Calculate discriminant $D$: $D = b^2 - 4ac = (-2\sqrt{2})^2 - 4(2)(1) = (4 \times 2) - 8 = 8 - 8 = 0$
3. Since $D = 0$, the equation has two equal real roots: $x = \frac{-b \pm \sqrt{0}}{2a} = \frac{-(-2\sqrt{2})}{2(2)} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$
4. Therefore, <u>the roots are $x = \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$</u>.

Solved Example 3: Fractional Form Equations

Problem: Find the roots of $x + \frac{1}{x} = 3$ ($x \ne 0$).

Solution:

1. Clear the fraction by multiplying throughout by $x$: $x^2 + 1 = 3x \implies x^2 - 3x + 1 = 0$
2. Here $a = 1, b = -3, c = 1$.
3. Compute discriminant: $D = b^2 - 4ac = (-3)^2 - 4(1)(1) = 9 - 4 = 5$
4. Since $D = 5 > 0$, real roots exist ($\\sqrt{D} = \sqrt{5}$ is irrational).
5. Apply the formula: $x = \frac{-(-3) \pm \sqrt{5}}{2(1)} = \frac{3 \pm \sqrt{5}}{2}$
6. Therefore, <u>the roots are $x = \frac{3 + \sqrt{5}}{2}$ and $x = \frac{3 - \sqrt{5}}{2}$</u>.

5. Summary and Examination Tips

Remember: In $-b$, if $b$ is already negative (e.g., $b = -5$), then $-b = -(-5) = +5$. Forgetting this double negative sign is the single most common student mistake in board exams!

Exam Tip: Always calculate $D = b^2 - 4ac$ first before substituting into the main formula. If $D < 0$, you save significant time because no further arithmetic is needed!