# Quadratics

## Solving Quadratics

A quadratic expression is any expression with an x squared term, an x term and a constant. For example, $ 3x^2-4x+7$ is a quadratic expression. Note that it doesn’t have to be an $x$, it could by $y$ or any other letter as long as it is the same throughout. Furthermore, there are three ways in which you can solve quadratics – each method requires setting the quadratic to 0 first. See Example 1.

### Factorising

Firstly, the simplest method, provided that it is possible, is factorising.

*Example: *$2x^2-5x-3=0\hspace{5pt}\Rightarrow\hspace{5pt}(2x+1)(x-3)=0\hspace{5pt}\Rightarrow\hspace{5pt} x=-\frac{1}{2}, x=3$.

### Quadratic Formula

If factorising doesn’t work but a quadratic does have roots, the quadratic formula will find them instead. Recall that the discriminant will tell you how many roots a quadratic has. See Discriminants page.

The quadratic formula says that if $ax^2+bx+c=0$ then the roots are given by:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

*Example*: In the following quadratic, $a=1$, $b=3$ and $c=-3$:

$x^2+3x-3=0\hspace{5pt}\Rightarrow\hspace{5pt}x=\frac{-3\pm\sqrt{3^2-4\times 1\times-3}}{2}\hspace{5pt}\Rightarrow\hspace{5pt}x=\frac{-3 +\sqrt{21}}{2}, \frac{-3-\sqrt{21}}{2}$

given exactly, i.e. not as a rounded decimal. Rounded to two decimal places using a calculator, the solutions are $x=0.79$ and $x=-3.79$. See Example 2.

### Completing the Square

Alternatively, another infallible method for finding roots (if a quadratic is solvable) is to complete the square. See Completing the Square page and Example 3.

*Example: *$x^2+6x+5=0\hspace{3pt}\hspace{5pt}\Rightarrow\hspace{5pt}(x+3)^2-4=0\hspace{5pt}\Rightarrow\hspace{5pt} x+3=\pm 2\hspace{3pt}\hspace{5pt}\Rightarrow\hspace{5pt} x=-1, x=-5$

## Sketching Quadratics

It is worth noting that completing the square is also useful for sketching a quadratic. The reason for this is that, by writing the quadratic in completed square form, we can see the transformations applied to the graph of $x^2$ (the shape of a quadratic is a known as a parabola). For example, to get $y=(x+3)^2+1$ we shift the graph of $x^2$ by 3 ($x$ transformation) and then up by 1 ($y$ transformation).

**1**. Firstly, find the **roots **using one of the above methods, roots occur when $y=0$.

**2.** Then, find the **$y$-intercept**, this occurs when $x=0$.

**3.** Finally, find the coordinates of the **vertex **by completing the square and applying transformations to $y=x^2$.

See Completing the Square for more details and check out Example 4.

**DESMOS **is a fantastic sketching tool. Click here to try it out. Firstly click the start graphing button and type y=x^2+4x-5 in the bar where the cursor starts. Then try adding more graphs and experimenting with the options. Finally, try exporting your graphs.

## Examples of Quadratics

Solve the following:

- $x^2+x=12$
- $2x^2=12x-16$

- Write $x^2+x=12$ as $x^2-x-12=0$ and then factorise to $(x-4)(x+3)=0$. The two factors $x-4$ and $x+3$ could both be zero to give a product of 0 and so the solutions are $x=4$ or $x=-3$.
- First write $2x^2=12x-16$ as $2x^2-12x+16=0$ and then divide both sides by 2 to give $x^2-6x+8=0$. Then factorise to $(x-4)(x-2)=0$ and so the solutions are $x=4$ and $x=2$.

Solve $2x^2-5x-6=0$, giving your answers to two decimal places.

Since the question says find the solutions to two decimal places, it suggests that we should use the quadratic formula. Selecting $a=2$, $b=-5$ and $c=-6$, the solutions are given by

$\begin{array}{c}x&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=&\frac{5\pm\sqrt{(-5)^2-4\times 2\times -6}}{2\times 2}\\&=&\frac{5\pm\sqrt{73}}{4}\\&\approx&3.39, -0.89\end{array}$

Solve the equation $2x^2+6x-9=0$ by completing the square. Give exact answers, simplified where appropriate.

Taking out a factor of 2 gives the equation:

$\begin{array}{c}2\left(x^2+3x-\frac{9}{2}\right)&=&0\\\Rightarrow \hspace{5pt}2\left(\left(x+\frac{3}{2}\right)^2-\frac{27}{4}\right)&=&0\\\Rightarrow \hspace{5pt}\left(x+\frac{3}{2}\right)^2&=&\frac{27}{4}\\\Rightarrow \hspace{5pt}x&=&-\frac{3}{2}\pm\sqrt{\frac{27}{4}}\end{array}$.

Hence, $x=-\frac{3}{2}+\frac{3\sqrt{3}}{2}$ and $x= -\frac{3}{2}-\frac{3\sqrt{3}}{2}$.

### Extra Notes/Questions

## Quadratics Videos

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