Solving Quadratics through Factorising

The quickest and easiest way to solve quadratic equations is by factorising.

You need to be able to spot ‘disguised‘ quadratics involving a function of x, f(x), instead of x itself. You need to use the substitution y=f(x) and solve for y, and then use these to find the values of x.

Example: Solve x^2 + 2x = 15 through factorising.

Rearrange the equation into the form ax^2 + bx + c = 0:

x^2 + 2x - 15 = 0

Then, solve the equation by factorising:

(x-3)(x+5) = 0

So,

x - 3 = 0 \Rightarrow \textcolor{blue}{x = 3}  and  x + 5 = 0 \Rightarrow \textcolor{blue}{x = -5}

Completing the Square Method

The following method can be used to complete the square of a quadratic expression:

Step 1: Rearrange the quadratic in the form

ax^2 + bx + c

Step 2: Take out a factor of a out of the x^2 and x terms:

a \left( x^2 + \dfrac{b}{a} x \right) + c

Step 3: Halve the coefficient of x and rewrite the brackets as one squared bracket:

a \left(x + \textcolor{limegreen}{\dfrac{b}{2a}} \right)^2

Step 4: Add \textcolor{red}{d} to the bracket to complete the square:

a \left(x + \textcolor{limegreen}{\dfrac{b}{2a}} \right)^2 + \textcolor{red}{d}

Step 5: Find \textcolor{red}{d} by setting this expression equal to the original expression:

a \left(x + \textcolor{limegreen}{\dfrac{b}{2a}} \right)^2 + \textcolor{red}{d} = ax^2 + bx + c

Solving this gives

\textcolor{red}{d} = \left( c - \dfrac{b^2}{4a} \right)

Step 6: Put it all together:

a \left(x + \dfrac{b}{2a} \right)^2 + \left( c - \dfrac{b^2}{4a} \right) = ax^2 + bx + c

Completing the Square Formula

You do not need to do the full steps of working from above when complete the square of a quadratic, you can just express it in the form,

ax^2 + b x + c = a \left(x + \textcolor{limegreen}{e} \right)^2 + \textcolor{red}{d}

where

\textcolor{limegreen}{e} =\dfrac{b}{2a} \,\,\, and \,\, \textcolor{red}{d} = c-\dfrac{b^2}{4a} = c-a \textcolor{limegreen}{e}^2

Solving Quadratics through Completing the Square

Completing the square isn’t the easiest method to solve quadratics, but it is useful when finding exact solutions – i.e. solutions involving surds etc.

Example: Find the exact solutions to 2x^2 - 12x + 14 = 0 by completing the square.

So a = 2, b = -12 and c = 14.

Hence,

\begin{aligned} \textcolor{limegreen}{d} &= \dfrac{b}{2a} = \dfrac{-12}{2 \times 2} = \textcolor{limegreen}{-3} \\[1.2em] \textcolor{red}{e} &= c - ad^2 = 14 - 2 (-3)^2 = 14 - 18 = \textcolor{red}{-4} \end{aligned}

So,

2x^2 - 12x + 14 = 2(x \textcolor{limegreen}{-3})^2 \textcolor{red}{- 4}

Put this equal to 0 and solve for x:

\begin{aligned} 2(x-3)^2 - 4 &= 0 \\ 2(x-3)^2 &= 4 \\ (x-3)^2 &= 2 \\ x - 3 &= \pm \sqrt{2} \end{aligned}

Hence,

\textcolor{blue}{x = 3 + \sqrt{2}}  and  \textcolor{blue}{x = 3 - \sqrt{2}}