Solving Quadratic Equations by the Formulae

The method of completing the square can be best used to establish a formula for the solution of the quadratic equation of the form 

\( \scriptsize ax^2 \: +\: bx\: +\: c = 0 \)

The formula is given as:

x = \(\large \frac{-b \pm \sqrt {b^2 \: – \: 4ac}}{2a}\)

The formula above is known as the quadratic formula, where

a is the coefficient of \( \scriptsize x^2 \)

b = coefficient of x

c = constant term.

Note that, the quadratic formula is particularly used to solve quadratic equations, which cannot be solved using factorization method.

Example 7.2.1:

Solve the following equations and leave your answers in 2 decimal places (d.p) using the quadratic formula.

i. \( \scriptsize 3x^2 \: – \: 5x \: – \: 9 = 0 \)

ii. \( \scriptsize – 3x^2 = (x\:+\:2)(x\:-\:1) \)

iii. \( \scriptsize \left(\scriptsize 2x \:- \: 3 \right)^2 = x^2 \)

iv. \( \scriptsize -4x^2 \: + \: 2x \: + \: 6 = 0 \)

Solution:

i. \( \scriptsize 3x^2 \: – \: 5x \: – \: 9 = 0 \)

Let a = 3, b = -5 and c = -9

Using the factor formula we have:

x = \(\frac{-b \pm \sqrt {b^2 \: – \: 4ac}}{2a}\)

x = \(\frac{-(-5) \pm \sqrt {5^2 \: – \: 4 \: \times \: 3 \: \times \: (-9)}}{2 \: \times \: 3}\)

x = \(\frac{5 \pm \sqrt {25\: + \:108}}{6}\)

x = \(  \frac{5 \pm \sqrt{133} }{6}  \) 

x = \(  \frac{5 \pm 11.53 }{6}  \) 

x = \( \frac{16.53 }{6} \:or \:  \frac{-6.53}{6} \)

x = 2.76 or 1.09

ii. \( \scriptsize – 3x^2 = (x\:+\:2)(x\:-\:1) \)

Expand the brackets

⇒ \( \scriptsize – 3x^2 = x^2 \:+\:2x \:-\: x \: – \: 2 \)

Collect like terms

⇒ \( \scriptsize 4x^2 \: + \: x \: – \: 2 = 0 \)

a = 4,      b = 1,   c = – 2

Using the factor formula we have:

x = \( \normalsize \frac{-b \pm \sqrt {b^2 \: – \: 4ac}}{2a}\)

x = \(\frac{-1 \pm \sqrt {1^2 \: – \: 4 \: \times \: 4 \: \times \: (-2)}}{2 \: \times \: 4}\)

x = \(\frac{-1 \pm \sqrt {1 \:+ \:32}}{8}\)

x = \(  \frac{-1 \pm \sqrt{33} }{8}  \) 

x = \(  \frac{-1 \pm 5.74 }{8}  \) 

x = \( \frac{-1 \:- \: 5.74 }{8} \:or \:  \frac{-1\:+ \: 5.74 }{8} \)

x = \( \frac{-6.74 }{8} \:or \:  \frac{4.74}{8} \)

x = – 0.84 or 0.59

iii. \( \scriptsize \left(\scriptsize 2x \:- \: 3 \right)^2 = x^2 \)

Expand the brackets

⇒ \( \scriptsize \left(\scriptsize 2x \:- \: 3 \right) \left(\scriptsize 2x \:- \: 3 \right) = x^2 \)

\( \scriptsize 4x^2 \: – \: 6x\: – \: 6x \: + \: 9 = x^2 \)

Collect like terms

⇒ \( \scriptsize 3x^2 \: – \:12x \: + \: 9 = 0 \)

Divide both sides by 3

⇒ \( \scriptsize x^2 \: – \:4x \: + \: 3 = 0 \)

Using the quadratic formula

a = 1,     b = -4,     c = 3

x = \( \normalsize \frac{-b \pm \sqrt {b^2 \: – \: 4ac}}{2a}\)

x = \(\frac{-(-4) \pm \sqrt {-4^2 \: – \: 4 \: \times \: 1 \: \times \: 3}}{2 \: \times \: 1}\)

x = \(\frac{4 \pm \sqrt {16 \: – \: 12}}{2}\)

x = \(\frac{4 \pm \sqrt {4}}{2}\)

x = \(  \frac{4 \pm 2}{2}  \) 

x = \( \frac{4 + 2 }{2} \:or \:  \frac{4 \: – \: 2}{2} \)

x = \( \frac{6 }{2} \:or \: \frac{2}{2} \)

x = 3 or 1

Alternate Method

\( \scriptsize \left(\scriptsize 2x \:- \: 3 \right)^2 = x^2 \)

Take the square root of both sides

⇒      x = 2x – 3

⇒      x = 3

Note: This method only provides one solution.

iv. \( \scriptsize -4x^2 \: + \: 2x \: + \: 6 = 0 \)

Divide both sides by 2

⇒ \( \scriptsize -2x^2 \: + \: x \: + \: 3 = 0 \)

Re-arrange the terms

⇒ \( \scriptsize 2x^2 \: – \: x \: – \: 3 = 0 \)

Using the quadratic formula

⇒ a = 2,       b = -1      and    c = -3

x = \( \normalsize \frac{-b \pm \sqrt {b^2 \: – \: 4ac}}{2a}\)

x = \(\frac{-(-1) \pm \sqrt {(-1)^2 \: – \: 4 \: \times \:2 \: \times \: (-3)}}{2 \: \times \: 2}\)

x = \(\frac{1 \pm \sqrt {25}}{4}\)

x = \(  \frac{1 \pm 5}{4}  \) 

x = \( \frac{1 \:+\: 5 }{4} \:or \:  \frac{1 \: – \: 5}{2} \)

x = \( \frac{6 }{4} \:or \: \frac{-4}{4} \)

x = 1.5 or – 1