Quadratic Equations

ICSE Class 10 Mathematics • Chapter 04

1. Definition and Standard Form

Quadratic Equation: A polynomial equation of degree 2 in one variable.

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

Roots: Values of $x$ that satisfy the equation (also called solutions or zeros).

2. Methods of Solving

Method 1: Factorisation (Splitting the Middle Term)

Steps for $ax^2 + bx + c = 0$:
  1. Find product: $a \times c$
  2. Find two numbers whose product = ac and sum = b
  3. Split the middle term $bx$ using these numbers
  4. Take common factors and factorise
  5. Set each factor = 0 and solve

Example 1: Solve $x^2 - 7x + 12 = 0$ by factorisation.

Solution:

Step Working
1. Find ac $1 \times 12 = 12$
2. Find factors Product = 12, Sum = −7
Numbers: −3 and −4
3. Split middle term $x^2 - 3x - 4x + 12 = 0$
4. Factor by grouping $x(x - 3) - 4(x - 3) = 0$
$(x - 3)(x - 4) = 0$
5. Solve $x - 3 = 0$ or $x - 4 = 0$
$x = 3$ or $x = 4$

Verification: Sum of roots = 3 + 4 = 7 = $-\frac{-7}{1}$ ✓

Product of roots = 3 × 4 = 12 = $\frac{12}{1}$ ✓

Method 2: Quadratic Formula (Shreedharacharya's Rule)

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

This always works, even when factorisation is difficult!

Steps:
  1. Identify a, b, c from standard form
  2. Calculate discriminant: $D = b^2 - 4ac$
  3. Substitute in formula
  4. Two roots: $x = \frac{-b + \sqrt{D}}{2a}$ and $x = \frac{-b - \sqrt{D}}{2a}$

Example 2: Solve $2x^2 + 5x - 3 = 0$ using the quadratic formula.

Solution:

$a = 2$, $b = 5$, $c = -3$

$D = b^2 - 4ac = 25 - 4(2)(-3) = 25 + 24 = 49$

$x = \frac{-5 \pm \sqrt{49}}{2(2)} = \frac{-5 \pm 7}{4}$

$x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}$ or $x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3$

Roots: $x = \frac{1}{2}$ or $x = -3$

Method 3: Completing the Square

Steps for $ax^2 + bx + c = 0$:
  1. Divide by $a$ to make coefficient of $x^2$ equal to 1
  2. Move constant to RHS: $x^2 + \frac{b}{a}x = -\frac{c}{a}$
  3. Add $\left(\frac{b}{2a}\right)^2$ to both sides
  4. LHS becomes perfect square: $\left(x + \frac{b}{2a}\right)^2$
  5. Take square root and solve

3. Discriminant and Nature of Roots

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

Value of D Nature of Roots Graph Interpretation
$D > 0$ Two distinct real roots Parabola cuts x-axis at 2 points
$D = 0$ Two equal (repeated) real roots Parabola touches x-axis at 1 point
$D < 0$ No real roots (imaginary) Parabola doesn't touch x-axis

ICSE Exam Favourite: "Find value of k for which roots are equal" → Use $D = 0$

Example 3: Find value of $k$ for which $x^2 - 4x + k = 0$ has equal roots.

Solution:

For equal roots: $D = 0$

$b^2 - 4ac = 0$

$(-4)^2 - 4(1)(k) = 0$

$16 - 4k = 0$

$k = 4$

4. Sum and Product of Roots

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

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

Product of Roots: $\alpha \cdot \beta = \frac{c}{a}$

Forming Equation from Roots

If roots are $\alpha$ and $\beta$:

$x^2 - (\alpha + \beta)x + \alpha\beta = 0$

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

Example 4: If roots of $2x^2 - 5x + 3 = 0$ are $\alpha, \beta$, find:

(a) $\alpha + \beta$ (b) $\alpha\beta$ (c) $\alpha^2 + \beta^2$ (d) $\frac{1}{\alpha} + \frac{1}{\beta}$

Solution:

(a) $\alpha + \beta = -\frac{-5}{2} = \frac{5}{2}$

(b) $\alpha\beta = \frac{3}{2}$

(c) $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{25}{4} - 3 = \frac{25-12}{4} = \frac{13}{4}$

(d) $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5/2}{3/2} = \frac{5}{3}$

5. Useful Identities for Sum/Product Problems

Expression In Terms of Sum & Product
$\alpha^2 + \beta^2$ $(\alpha + \beta)^2 - 2\alpha\beta$
$\alpha^2 - \beta^2$ $(\alpha + \beta)(\alpha - \beta)$
$(\alpha - \beta)^2$ $(\alpha + \beta)^2 - 4\alpha\beta$
$\frac{1}{\alpha} + \frac{1}{\beta}$ $\frac{\alpha + \beta}{\alpha\beta}$
$\alpha^3 + \beta^3$ $(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$

6. Word Problems

Example 5: The product of two consecutive positive integers is 306. Find them.

Solution:

Let integers be $n$ and $n+1$

$n(n+1) = 306$

$n^2 + n - 306 = 0$

Using formula or factorisation: $(n-17)(n+18) = 0$

$n = 17$ (rejecting -18 as we need positive)

Integers: 17 and 18

Example 6: A train travels 360 km at uniform speed. If speed had been 5 km/h more, it would have taken 1 hour less. Find the speed.

Solution:

Let speed = $v$ km/h, Time = $\frac{360}{v}$ hours

New speed = $(v+5)$ km/h, New time = $\frac{360}{v+5}$ hours

$\frac{360}{v} - \frac{360}{v+5} = 1$

$360(v+5) - 360v = v(v+5)$

$1800 = v^2 + 5v$

$v^2 + 5v - 1800 = 0$

$(v+45)(v-40) = 0$

Speed = 40 km/h

7. Quick Reference

Formula/Concept Expression
Standard Form $ax^2 + bx + c = 0$
Quadratic Formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
Discriminant $D = b^2 - 4ac$
Sum of Roots $-\frac{b}{a}$
Product of Roots $\frac{c}{a}$
Equation from Roots $x^2 - (\text{Sum})x + (\text{Product}) = 0$

Exam Practice (PYQ Trends)

PYQ: 2023

BOARD Solve for $x$: $\frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4}$

PYQ: 2022

BOARD The sum of two numbers is 15 and the sum of their reciprocals is $\frac{3}{10}$. Find the numbers.