Chapter 5: Quadratic Equations

A quadratic equation is a polynomial equation of second degree in one variable. In this chapter, we master standard representation, the discriminant test for nature of roots, solving methods (factorisation & quadratic formula), and solving equations reducible to quadratic form.

1. Definition & Standard Form

A polynomial equation in one variable $x$ in which the highest power of $x$ is $2$ is called a Quadratic Equation.

Standard Form

The standard form of a quadratic equation is:

$$ax^2 + bx + c = 0 \quad \text{where } a, b, c \in \mathbb{R} \text{ and } a \neq 0$$

$a$ = coefficient of $x^2$ (leading coefficient; $a \neq 0$)
$b$ = coefficient of $x$
$c$ = constant term

Classification

Affected Quadratic Equation: Contains both $x^2$ and $x$ terms (i.e. $b \neq 0$). E.g., $3x^2 + 5x - 2 = 0$.
Pure Quadratic Equation: Contains only the $x^2$ term (i.e. $b = 0$). E.g., $4x^2 - 9 = 0$.

Roots (Solutions): The values of $x$ which satisfy $ax^2 + bx + c = 0$ (make $\text{LHS} = \text{RHS} = 0$). Every quadratic equation has exactly two roots.

2. Discriminant ($D$) & Nature of Roots

The expression $b^2 - 4ac$ is called the Discriminant of the quadratic equation $ax^2 + bx + c = 0$, denoted by $D$.

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

Value of Discriminant ($D$)	Nature of Roots	Form of Roots
$D > 0$ (Perfect Square)	Real, Unequal (Distinct) & Rational	$x = \frac{-b \pm \sqrt{D}}{2a}$
$D > 0$ (Not a Perfect Square)	Real, Unequal (Distinct) & Irrational	$x = \frac{-b \pm \sqrt{D}}{2a}$ (Surd pairs)
$D = 0$	Real & Equal (Repeated / Coincident)	$x = -\frac{b}{2a}, -\frac{b}{2a}$
$D < 0$	No Real Roots (Imaginary)	$\sqrt{-D}$ is non-real

ICSE Key Tip

For questions asking to "Find $k$ for which roots are equal", set $D = b^2 - 4ac = 0$ and solve for $k$.

Practice Problems — Discriminant & Nature of Roots

Q1. Without solving, examine the nature of roots of $3x^2 - 4\sqrt{3}x + 4 = 0$.

Solution:
$a = 3, b = -4\sqrt{3}, c = 4$
$D = b^2 - 4ac = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$.
Since $D = 0$, the roots are real and equal.
Ans: Real & Equal Roots

Q2. Find the value of $m$ if $(4+m)x^2 + (m+1)x + 1 = 0$ has equal roots.

Solution:
$a = 4+m, b = m+1, c = 1$
For equal roots, $D = b^2 - 4ac = 0 \Rightarrow (m+1)^2 - 4(4+m)(1) = 0$
$m^2 + 2m + 1 - 16 - 4m = 0 \Rightarrow m^2 - 2m - 15 = 0$
$(m-5)(m+3) = 0 \Rightarrow m = 5$ or $m = -3$.
Ans: m = 5 or m = -3

3. Methods of Solving Quadratic Equations

Method A: Factorisation (Splitting Middle Term)

Zero Product Rule

If $A \times B = 0$, then either $A = 0$ or $B = 0$.

To split middle term $bx$ in $ax^2 + bx + c = 0$, find numbers $p$ and $q$ such that $p \times q = a \times c$ and $p + q = b$.

Example: Solve $\sqrt{3}x^2 + 11x + 6\sqrt{3} = 0$.

Solution:
$a = \sqrt{3}, c = 6\sqrt{3} \Rightarrow ac = \sqrt{3} \times 6\sqrt{3} = 18$.
Find factors of 18 adding to 11 $\Rightarrow 9$ and $2$.
$\sqrt{3}x^2 + 9x + 2x + 6\sqrt{3} = 0 \Rightarrow \sqrt{3}x(x + 3\sqrt{3}) + 2(x + 3\sqrt{3}) = 0$
$(x + 3\sqrt{3})(\sqrt{3}x + 2) = 0 \Rightarrow x = -3\sqrt{3}$ or $x = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$.
Ans: x = -3√3 or x = -2√3/3

Method B: Quadratic Formula (Shreedharacharya's Rule)

Derived by completing the square on $ax^2 + bx + c = 0$:

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

Practice Problems — Quadratic Formula & Decimal Accuracy

Q1. Solve $x - \frac{18}{x} = 6$. Give answer correct to 2 significant figures. [ICSE]

Solution:
Multiply by $x$: $x^2 - 18 = 6x \Rightarrow x^2 - 6x - 18 = 0$
$a = 1, b = -6, c = -18$
$D = (-6)^2 - 4(1)(-18) = 36 + 72 = 108$
$\sqrt{108} \approx 10.3923$
$x = \frac{6 \pm 10.3923}{2} \Rightarrow x_1 = \frac{16.3923}{2} = 8.196 \approx 8.2$, $x_2 = \frac{-4.3923}{2} = -2.196 \approx -2.2$.
Ans: x = 8.2 or x = -2.2

4. Equations Reducible to Quadratic Form

Substitution Strategies

Biquadratic ($ax^4 + bx^2 + c = 0$): Substitute $x^2 = y$.
Repeated Bracket [$a(x^2+3x)^2 + b(x^2+3x) + c = 0$]: Substitute $x^2 + 3x = y$.
Reciprocal Radicals [$\sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}} = k$]: Substitute $\sqrt{\frac{x}{1-x}} = y \Rightarrow y + \frac{1}{y} = k$.

Example: Solve $\sqrt{\frac{x}{x-3}} + \sqrt{\frac{x-3}{x}} = \frac{5}{2}$.

Solution:
Let $y = \sqrt{\frac{x}{x-3}} \Rightarrow y + \frac{1}{y} = \frac{5}{2} \Rightarrow 2y^2 - 5y + 2 = 0 \Rightarrow (2y-1)(y-2) = 0$.
If $y = 2 \Rightarrow \frac{x}{x-3} = 4 \Rightarrow x = 4x - 12 \Rightarrow 3x = 12 \Rightarrow x = 4$.
If $y = \frac{1}{2} \Rightarrow \frac{x}{x-3} = \frac{1}{4} \Rightarrow 4x = x - 3 \Rightarrow 3x = -3 \Rightarrow x = -1$.
Ans: x = 4 or x = -1

5. Sum & Product of Roots

If $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$:

$$\text{Sum of Roots: } \alpha + \beta = -\frac{b}{a}$$
$$\text{Product of Roots: } \alpha\beta = \frac{c}{a}$$

Forming Quadratic Equation from Roots ($\alpha, \beta$):

$$x^2 - (\alpha + \beta)x + \alpha\beta = 0 \quad \text{i.e.} \quad x^2 - (\text{Sum})x + (\text{Product}) = 0$$

Summary Table

Concept	Mathematical Formula / Expression	Key Condition
Standard Form	$ax^2 + bx + c = 0$	$a \neq 0, a,b,c \in \mathbb{R}$
Discriminant ($D$)	$D = b^2 - 4ac$	$D > 0 \Rightarrow \text{Real/Distinct}; D = 0 \Rightarrow \text{Equal}$
Quadratic Formula	$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$	Applies to all quadratic equations
Sum of Roots ($\alpha+\beta$)	$-\frac{b}{a}$	$\alpha, \beta$ are roots
Product of Roots ($\alpha\beta$)	$\frac{c}{a}$	$\alpha, \beta$ are roots
Forming Equation	$x^2 - (\text{Sum})x + (\text{Product}) = 0$	Given roots $\alpha, \beta$

Tips for ICSE Board Exams

Check if the question asks for decimal places (e.g. 2 decimal places) or significant figures. Calculate square root values accurately to 3 places before rounding off.
In equal root problems ($D = 0$), don't forget to check if $a \neq 0$ for the calculated values of parameters like $k$ or $m$.
When using substitutions ($y = x^2$ or $y = x + 1/x$), remember to convert back to find $x$.
Always verify roots by substituting back into the original equation whenever time permits.