Solving Quadratic Equations

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  $textcolor{blue}{x = 3 – sqrt{2}}$