Quadratic Equations

A quadratic equation is a polynomial equation of degree 2 in the variable \(x\), expressed in the standard form:

\[ ax^2 + bx + c = 0, \quad a \neq 0 \]

where \(a\), \(b\), and \(c\) are real numbers. The solutions or roots of this equation are the values of \(x\) that satisfy the equation.

Formula Derivation: Quadratic Formula

To find the roots of \(ax^2 + bx + c = 0\), we complete the square as follows:

Start with:

\[ ax^2 + bx + c = 0 \]

Divide both sides by \(a\) (since \(a \neq 0\)):

\[ x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \]

Rearranged:

\[ x^2 + \frac{b}{a}x = -\frac{c}{a} \]

Add \(\left(\frac{b}{2a}\right)^2\) to both sides to complete the square:

\[ x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2 \]

Left side becomes a perfect square:

\[ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \]

Taking square roots:

\[ x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} \]

Therefore,

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

Worked Illustrations and Solved Examples

Example 1: Check if \((x - 2)^2 + 1 = 2x - 3\) is a quadratic equation.

Solution:

Expand and simplify:

\[ (x - 2)^2 + 1 = 2x - 3 \]

\[ x^2 - 4x + 4 + 1 = 2x - 3 \]

\[ x^2 - 4x + 5 = 2x - 3 \]

Bring all terms to one side:

\[ x^2 - 4x + 5 - 2x + 3 = 0 \]

\[ x^2 - 6x + 8 = 0 \]

This is in the form \(ax^2 + bx + c = 0\) with \(a=1, b=-6, c=8\), so it is a quadratic equation.

Example 2: Find roots of \(2x^2 - 5x + 3 = 0\) by factorisation.

Solution:

Multiply \(a\) and \(c\): \(2 \times 3 = 6\).

Find two numbers whose product is 6 and sum is \(-5\): \(-2\) and \(-3\).

Rewrite middle term:

\[ 2x^2 - 2x - 3x + 3 = 0 \]

Group terms:

\[ 2x(x - 1) - 3(x - 1) = 0 \]

Factor:

\[ (2x - 3)(x - 1) = 0 \]

Set each factor to zero:

\[ 2x - 3 = 0 \Rightarrow x = \frac{3}{2} \]

\[ x - 1 = 0 \Rightarrow x = 1 \]

Roots are \(1\) and \(\frac{3}{2}\).

Example 3: Find roots of \(6x^2 - x - 2 = 0\) using quadratic formula.

Solution:

Identify coefficients: \(a=6, b=-1, c=-2\).

Calculate discriminant:

\[ D = b^2 - 4ac = (-1)^2 - 4(6)(-2) = 1 + 48 = 49 \]

Apply quadratic formula:

\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{1 \pm 7}{12} \]

Roots:

\[ x = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3} \]

\[ x = \frac{1 - 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]

Practice Set

• Level 1 – Easy
  • Check if \(x^2 + 5x + 6 = 0\) is a quadratic equation.
  • Find roots of \(x^2 - 7x + 10 = 0\) by factorisation.
  • Find roots of \(x^2 + 4x + 4 = 0\) using quadratic formula.
• Level 2 – Moderate
  • Find roots of \(3x^2 - 2x - 1 = 0\) using quadratic formula.
  • Determine the nature of roots of \(2x^2 + 3x + 5 = 0\) using discriminant.
  • Find roots of \(4x^2 - 12x + 9 = 0\) by factorisation.
• Level 3 – Challenging
  • Find the roots of \(5x^2 + 6x + 1 = 0\) using quadratic formula and verify by factorisation.
  • For the quadratic equation \(ax^2 + bx + c = 0\), prove that sum of roots is \(-\frac{b}{a}\) and product of roots is \(\frac{c}{a}\).
  • Apply quadratic equations to solve: The area of a rectangle is 48 sq. units and its length is 3 units more than its width. Find the dimensions.

Answer Key

• Level 1
  • Yes, it is quadratic with \(a=1, b=5, c=6\).
  • Roots: \(x= -2, -3\).
  • Roots: \(x= -2, -2\) (repeated root).
• Level 2
  • Roots: \(x=1, -\frac{1}{3}\).
  • Discriminant \(D = 3^2 - 4(2)(5) = 9 - 40 = -31 < 0\), no real roots.
  • Roots: \(x= \frac{3}{2}, \frac{3}{2}\) (repeated root).
• Level 3
  • Quadratic formula roots: \(x= -\frac{1}{5}, -1\). Factorisation: \((5x + 1)(x + 1) = 0\) roots match.
  • Proof:
    • Sum of roots \(\alpha + \beta = -\frac{b}{a}\) from quadratic formula.
    • Product of roots \(\alpha \beta = \frac{c}{a}\).
  • Let width = \(x\), length = \(x + 3\).

  Area equation: \(x(x + 3) = 48\)

  \(x^2 + 3x - 48 = 0\)

  Using quadratic formula:

  \[ x = \frac{-3 \pm \sqrt{9 + 192}}{2} = \frac{-3 \pm 15}{2} \]

  Positive root: \(x = 6\), length = \(9\).

Quick Reference

Concept	Formula/Result
Standard form	\(ax^2 + bx + c = 0, a \neq 0\)
Quadratic formula	\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Discriminant	\(D = b^2 - 4ac\)
Nature of roots	\(D > 0\): two distinct real roots \(D = 0\): two equal real roots \(D < 0\): no real roots
Sum of roots	\(-\frac{b}{a}\)
Product of roots	\(\frac{c}{a}\)

Glossary

• Quadratic Equation: Polynomial equation of degree 2.
• Roots/Solutions: Values of \(x\) satisfying the quadratic equation.
• Discriminant: Expression \(b^2 - 4ac\) determining nature of roots.
• Factorisation: Expressing quadratic as product of two linear factors.
• Quadratic Formula: Formula to find roots of quadratic equations.
• Sum of Roots: Sum of the solutions of the quadratic equation.
• Product of Roots: Product of the solutions of the quadratic equation.