Unit 0 - Required Prior Knowledge
Required Prior Knowledge
Questions
Solve these linear equations
a) \(x+3=0\)
b) \(x-7=0\)
c) \(2x-4=0\)
d) \(3x+6=0\)
e) \(2x-5=0\)
f) \(5x+7=0\)
g) \(7x+2=0\)
h) \(3-8x=0\)
Solutions
Get Ready
Questions
Solve the following quadratic equations by factorising
a) \(x^{2}-9x+14=0\)
b) \(x^{2}+10x+25=0\)
c) \(x^{2}+6x=0\)
d) \(2x^{2}+3x=20\)
Solutions
Notes
To solve a quadratic equation by factorising, you use the Null Factor Law.
This states that if \(a\times b=0\) then either \(a=0\) or \(b=0\).
This is usually the best method to try first in non-calculator questions.
Examples and Your Turns
Example
Solve$$\left(x-3\right)\left(x+2\right)=0$$
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$$\begin{matrix}x-3=0&,&x-2=0\\x=3&,&x=2\end{matrix}$$
Your Turn
Solve$$\left(x+4\right)\left(x-1\right)=0$$
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$$\begin{matrix}x+4=0&,&x-1=0\\x=-4&,&x=1\end{matrix}$$
Your Turn
Solve$$\left(2x+5\right)\left(x-3\right)=0$$
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$$\begin{matrix}2x+5=0&,&x-3=0\\x=-\frac{5}{2}&,&x=3\end{matrix}$$
Your Turn
Solve$$\left(3x-5\right)\left(5x+2\right)=0$$
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$$\begin{matrix}3x-5=0&,&5x+2=0\\x=\frac{5}{3}&,&x=-\frac{2}{5}\end{matrix}$$
Example
Solve$$x^{2}-x-6=0$$
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$$\left(x-3\right)\left(x+2\right)=0\\\begin{matrix}x-3=0&,&x+2=0\\x=3&,&x=-2\end{matrix}$$
Your Turn
Solve$$x^{2}+x-6=0$$
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$$\left(x+3\right)\left(x-2\right)=0\\\begin{matrix}x+3=0&,&x-2=0\\x=-3&,&x=2\end{matrix}$$
Your Turn
Solve$$x^{2}+3x-10=0$$
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$$\left(x+5\right)\left(x-2\right)=0\\\begin{matrix}x+5=0&,&x-2=0\\x=-5&,&x=2\end{matrix}$$
Your Turn
Solve$$x^{2}-7x=0$$
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$$x\left(x-7\right)=0\\\begin{matrix}x=0&,&x-7=0\\x=0&,&x=7\end{matrix}$$
Notes
Completing the Square is the process of converting a quadratic from the form \(ax^{2}+bx+c\) into the form \(a\left(x-h\right)^{2}+k\).
Thinking backwards, when we expand \(\left(x+m\right)^{2}\) we get:
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$$\left(x+m\right)^{2}=x^{2}+2mx+m^{2}$$
This means that when we complete the square we need the value in the brackets to be HALF the value of \(b\).
Examples and Your Turns
Example
Write \(x^{2}+4x+3\) in the form \(\left(x-h\right)^{2}+k\). Hence solve the equation \(x^{2}+4x+3=0\) exactly.
Your Turn
Write \(x^{2}-6x+5\) in the form \(\left(x-h\right)^{2}+k\). Hence solve the equation \(x^{2}-6x+5=0\) exactly.
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$$\begin{align}x^{2}-6x+5&=\left(x-3\right)^{2}-\left(3\right)^{2}+5\\&=\left(x-3\right)^{2}-4\end{align}$$Hence$$\left(x-3\right)^{2}-4=0\\\left(x-3\right)^{2}=4\\x-3=\pm 2\\x=3\pm 2\\x=5, x=1$$
Example
Write \(x^{2}+5x+16\) in the form \(\left(x-h\right)^{2}+k\). Hence solve the equation \(x^{2}+5x+16=0\) exactly.
Your Turn
Write \(x^{2}-x+7\) in the form \(\left(x-h\right)^{2}+k\). Hence solve the equation \(x^{2}-x+7=0\) exactly.
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$$\begin{align}x^{2}-x+7&=\left(x-\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+7\\&=\left(x-\frac{1}{2}\right)^{2}+\frac{27}{4}\end{align}$$Hence$$\left(x-\frac{1}{2}\right)^{2}+\frac{27}{4}=0\\ \left(x-\frac{1}{2}\right)^{2}=-\frac{27}{4}$$which is not possible, so there are no solutions.
Example
Write \(3x^{2}-48x+28\) in the form \(\left(x-h\right)^{2}+k\). Hence solve the equation \(3x^{2}-48x+28=0\) exactly.
Your Turn
Write \(2x^{2}+16x-10\) in the form \(\left(x-h\right)^{2}+k\). Hence solve the equation \(2x^{2}+16x-10=0\) exactly.
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$$\begin{align}2x^{2}+16x-10&=2\left(x^{2}+8x\right)-10\\&=2\left(\left(x+4\right)^{2}-\left(4\right)^{2}\right)-10\\&=2\left(x+4\right)^{2}-32-10\\&=2\left(x+4\right)^{2}-42\end{align}$$Hence$$2\left(x+4\right)^{2}-42=0\\ 2\left(x+4\right)^{2}=42\\ \left(x+4\right)^{2}=21\\x+4=\pm \sqrt{21}\\x=-4\pm \sqrt{21}$$
Example
Write \(2x^{2}+6x+18\) in the form \(\left(x-h\right)^{2}+k\). Hence solve the equation \(2x^{2}+6x+18=0\) exactly.
Your Turn
Write \(3x^{2}-4x+1\) in the form \(\left(x-h\right)^{2}+k\). Hence solve the equation \(3x^{2}-4x+1=0\) exactly.
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$$\begin{align}3x^{2}-4x+1&=3\left(x^{2}-\frac{4}{3}x\right)+1\\&=3\left(\left(x-\frac{2}{3}\right)^{2}-\left(\frac{2}{3}\right)^{2}\right)+1\\&=3\left(x-\frac{2}{3}\right)^{2}-\frac{4}{3}+1\\&=3\left(x-\frac{2}{3}\right)^{2}-\frac{1}{3}\end{align}$$Hence$$3\left(x-\frac{2}{3}\right)^{2}-\frac{1}{3}=0\\3\left(x-\frac{2}{3}\right)^{2}=\frac{1}{3}\\ \left(x-\frac{2}{3}\right)^{2}=\frac{1}{9}\\x-\frac{2}{3}=\pm \frac{1}{3}\\x=\frac{2}{3}\pm \frac{1}{3}\\x=1, x=\frac{1}{3}$$
Notes
The quadratic formula can be used to solve quadratic equations that cannot be factorised, to obtain exact solutions or those rounded to 3 significant figures.
The solutions of the equation \(ax^{2}+bx+c=0\) are given by$$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$$
Examples and Your Turns
Example
Solve \(2x^{2}+19x+45=0\), giving your answers exactly.
Your Turn
Solve \(2x^{2}+3x-6=0\), giving your answers exactly.
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We see that $$a=2, b=3, c=-6$$and substituting these into the quadratic formula gives$$\begin{align}x&=\frac{-3\pm\sqrt{3^{2}-4\left(2\right)\left(-6\right)}}{2\times 2}\\&=\frac{-3\pm\sqrt{57}}{4}\end{align}$$
Your Turn
By completing the square for \(ax^{2}+bx+c\), prove that the solutions to \(ax^{2}+bx+c=0\) are given by the quadratic formula.
Key Facts
Use this applet to generate a prompt for a Key Fact that you need to know for the course. The idea is that you should KNOW these key facts in order to be able to solve problems.