Quadratic Equations: Complete Discriminant & Formula Blueprint Class 10

In pure algebra, linear equations can only describe straight lines. But the physical universe is non-linear: gravitational acceleration accelerates falling bodies quadratically ($s = ut + \frac{1}{2}at^2$), kinetic energy scales with the square of velocity ($E_k = \frac{1}{2}mv^2$), and planar areas expand quadratically with side dimensions.

In CBSE Class 10 Mathematics, Chapter 4 (Quadratic Equations) represents a heavyweight topic in Section C and Section D. Many students struggle when confronted with complex, non-standard quadratics featuring algebraic fractions like $\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}$ or symbolic coefficients like $9x^2 - 9(a+b)x + (2a^2 + 5ab + 2b^2) = 0$.

In this master guide, we derive the Quadratic Formula (Sridharacharya's Rule) from first principles and conquer the most challenging quadratic equation problems on the board exam.

What You Will Learn

• Derivation of the Quadratic Formula by Completing the Square
• The Discriminant ($D = b^2 - 4ac$) and root nature classification
• Master Problem 1: Solving non-standard Fractional Quadratic Equations
• Master Problem 2: Solving Symbolic Literal Coefficients ($9x^2 - 9(a+b)x + \dots = 0$)
• Master Problem 3: Radical Quadratic Equations ($\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0$)
• Geometric word problems: The Rectangular Mango Grove and Perimeter problems
• Examiner presentation standards and arithmetic safeguards

1. Derivation of the Quadratic Formula (Completing the Square)

Let the general quadratic equation be: $ax^2 + bx + c = 0 \quad (a \ne 0)$

Step-by-Step Derivation:

1. Divide the entire equation by $a$: $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$
2. Move the constant term $\frac{c}{a}$ to the right-hand side: $x^2 + \frac{b}{a}x = -\frac{c}{a}$
3. Add the square of half of the coefficient of $x$—which is $\left(\frac{b}{2a}\right)^2$—to both sides: $x^2 + 2 \left(\frac{b}{2a}\right)x + \left(\frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}$
4. Express the left side as a perfect square $(x + \frac{b}{2a})^2$: $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}$
5. Take the square root of both sides: $x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$
6. Transpose $\frac{b}{2a}$ to the right-hand side: $\mathbf{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

The Sridharacharya Quadratic Formula is proved.

2. Master Problem 1: Fractional Quadratic Equation (NCERT Classic)

Problem Statement:

Find the roots of the equation: $\frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}, \quad x \ne -4, 7$

Step-by-Step Solution:

1. Combine the Left-Hand Side under a Common Denominator: $\frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30}$ $\frac{x - 7 - x - 4}{x^2 - 7x + 4x - 28} = \frac{11}{30}$ $\frac{-11}{x^2 - 3x - 28} = \frac{11}{30}$
2. Divide Both Sides by 11: $\frac{-1}{x^2 - 3x - 28} = \frac{1}{30}$
3. Cross-Multiply: $-(30) = x^2 - 3x - 28$ $-30 = x^2 - 3x - 28$ $x^2 - 3x - 28 + 30 = 0$ $\mathbf{x^2 - 3x + 2 = 0}$
4. Factorize by Splitting the Middle Term: Product $= 2$, Sum $= -3$ ($-2$ and $-1$): $(x - 2)(x - 1) = 0$ $\mathbf{x = 2} \quad \text{or} \quad \mathbf{x = 1}$
5. Since neither $1$ nor $2$ equals $-4$ or $7$, both solutions are valid.
6. Therefore, <u>the roots of the equation are $x = 1$ and $x = 2$</u>.

3. Master Problem 2: Radical Coefficients

Problem Statement:

Solve for $x$: $\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0$

Step-by-Step Solution:

1. Standard form: $a = \sqrt{3}, \quad b = 10, \quad c = 7\sqrt{3}$.
2. Product of $a$ and $c$: $a \times c = \sqrt{3} \times 7\sqrt{3} = 7 \times 3 = \mathbf{21}$
3. Look for two numbers whose product is $21$ and sum is $10$ ($7$ and $3$!): $\sqrt{3}x^2 + 3x + 7x + 7\sqrt{3} = 0$
4. Group terms and factor (Recall that $3 = \sqrt{3} \times \sqrt{3}$): $\sqrt{3}x(x + \sqrt{3}) + 7(x + \sqrt{3}) = 0$ $(x + \sqrt{3})(\sqrt{3}x + 7) = 0$
5. Solve for $x$: $x + \sqrt{3} = 0 \implies \mathbf{x = -\sqrt{3}}$ $\sqrt{3}x + 7 = 0 \implies \mathbf{x = -\frac{7}{\sqrt{3}} = -\frac{7\sqrt{3}}{3}}$
6. Therefore, <u>the roots are $x = -\sqrt{3}$ and $x = -\frac{7\sqrt{3}}{3}$</u>.

4. Master Problem 3: Symbolic Literal Coefficients (CBSE 4-Mark Heavyweight)

Problem Statement:

Solve for $x$: $9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2) = 0$

Step-by-Step Solution:

1. Factorize the Constant Term $(2a^2 + 5ab + 2b^2)$: $2a^2 + 4ab + ab + 2b^2 = 2a(a + 2b) + b(a + 2b) = \mathbf{(2a + b)(a + 2b)}$
2. Rewrite the Equation: $9x^2 - 9(a + b)x + (2a + b)(a + 2b) = 0$
3. Split the Middle Term $-9(a + b)$: Notice that: $3(2a + b) + 3(a + 2b) = 6a + 3b + 3a + 6b = 9a + 9b = 9(a + b)$
4. Rewrite the middle term: $9x^2 - 3(2a + b)x - 3(a + 2b)x + (2a + b)(a + 2b) = 0$
5. Factor by Grouping: $3x[3x - (2a + b)] - (a + 2b)[3x - (2a + b)] = 0$ $[3x - (2a + b)][3x - (a + 2b)] = 0$
6. Solve for $x$: $3x = 2a + b \implies \mathbf{x = \frac{2a + b}{3}}$ $3x = a + 2b \implies \mathbf{x = \frac{a + 2b}{3}}$
7. Therefore, <u>the roots are $x = \frac{2a + b}{3}$ and $x = \frac{a + 2b}{3}$</u>.

5. Summary and Examination Tips

Exam Tip: In the fractional problem $\frac{1}{x+4} - \frac{1}{x-7}$, always write the restriction $x \ne -4, 7$ in your first step! If one of your calculated roots turns out to be $-4$ or $7$, you must reject it because it causes division by zero.

Common Mistake: In the quadratic formula, forgetting that the denominator is $2a$. When $a \ne 1$, dividing only by $2$ instead of $2a$ invalidates the entire numerical answer!