Solving Quadratic Equations

Solving Quadratic Equations

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

Solving Quadratic Equations

You can solve quadratic equations of the form ax^2 + bx + c = 0 through factorising, completing the square or using the quadratic formula.

A Level AQA Edexcel OCR

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}

A LevelAQAEdexcelOCR

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

A LevelAQAEdexcelOCR

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

A LevelAQAEdexcelOCR

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}}

A LevelAQAEdexcelOCR

Note:

To solve some equations it may be easier to use the quadratic formula instead of factorising or completing the square, e.g. when the values of a, b and c are large. However the questions usually won’t tell you which method to use.

A Level AQA Edexcel OCR

Example: Solving ‘Disguised’ Quadratics

Solve x^4 - 13x^2 + 36 = 0

[3 marks]

f(x) = x^2, so let y = x^2 and substitute into the equation:

y^2 - 13y + 36 = 0

using x^4 = (x^2)^2 = y^2

Then, solve the quadratic by factorising:

(y-9)(y-4) = 0

y - 9 = 0 \Rightarrow \textcolor{purple}{y = 9}  and  y-4=0 \Rightarrow \textcolor{purple}{y = 4}

Use these y values to find the values of x:

y = 9 \Rightarrow x^2 = 9 \Rightarrow x = \sqrt{9} \Rightarrow \textcolor{blue}{x = \pm 3}

y = 4 \Rightarrow x^2 = 4 \Rightarrow x = \sqrt{4} \Rightarrow \textcolor{blue}{x = \pm 2}

A LevelAQAEdexcelOCR

Example Questions

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

2x^2 - 7x + 3 = 0

Then, solve the equation through factorising:

(2x-1)(x-3) = 0

So,

2x-1 = 0 \Rightarrow x = \dfrac{1}{2}  and  x-3 = 0 \Rightarrow x = 3

f(x) = x^{\frac{1}{2}} so let y = x^{\frac{1}{2}} and substitute into the equation:

 

y^2 - y - 6 = 0

using x = (x^{\frac{1}{2}})^2 = y^2

 

Then, solve the quadratic by factorising:

(y-3)(y+2) = 0

y - 3 = 0 \Rightarrow y = 3  and  y+2 = 0 \Rightarrow y = -2

 

Use these values of y to find the values of x:

y = 3 \Rightarrow x^{\frac{1}{2}} = 3 \Rightarrow x = 3^2 = 9

y = -2 \Rightarrow x^{\frac{1}{2}} = -2 \Rightarrow x = (-2)^2 = 4

a) a = 3, b = -30 and c = 73.

Hence,

d = \dfrac{b}{2a} = \dfrac{-30}{2 \times 3} = -5

and

e = c - ad^2 = 73 - 3(-5)^2 = 73 - 75 = -2

So,

3x^2 - 30x + 73 = 3(x-5)^2 - 2

 

b) 3(x-5)^2 - 2 = 0

 

So,

 

3(x-5)^2 = 2

 

(x-5)^2 = \dfrac{2}{3}

 

x - 5 = \pm \sqrt{\dfrac{2}{3}} = \dfrac{\sqrt{6}}{3}

 

x = 5 + \dfrac{\sqrt{6}}{3} = 5.82 (2 dp)

 

x = 5 - \dfrac{\sqrt{6}}{3} = 4.18 (2 dp)

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