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Equations and inequalities: solving quadratic equations

A quadratic equation is an equation where the power of the variable is at most 2. The following are examples of quadratic equations.

$\begin{array}{ccc}\hfill 2{x}^{2}+2x& =& 1\hfill \\ \hfill \frac{2-x}{3x+1}& =& 2x\hfill \\ \hfill \frac{4}{3}x-6& =& 7{x}^{2}+2\hfill \end{array}$

Quadratic equations differ from linear equations by the fact that a linear equation only has one solution, while a quadratic equation has at most two solutions. However, there are some special situations when a quadratic equation only has one solution.

We solve quadratic equations by factorisation, that is writing the quadratic as a product of two expressions in brackets. For example, we know that:

$\left(x+1\right)\left(2x-3\right)=2{x}^{2}-x-3.$

In order to solve:

$2{x}^{2}-x-3=0$

we need to be able to write $2{x}^{2}-x-3$ as $\left(x+1\right)\left(2x-3\right)$ , which we already know how to do. The reason for equating to zero and factoring is that if we attempt to solve it in a 'normal' way, we may miss one of the solutions. On the other hand, if we have the (non-linear) equation $f\left(x\right)g\left(x\right)=0$ , for some functions $f$ and $g$ , we know that the solution is $f\left(x\right)=0$ OR $g\left(x\right)=0$ , which allows us to find BOTH solutions (or know that there is only one solution if it turns out that $f=g$ ).

1. $x+{x}^{2}$
2. ${x}^{2}+1+2x$
3. ${x}^{2}-4x+5$
4. $16{x}^{2}-9$
5. $4{x}^{2}+4x+1$

Being able to factorise a quadratic means that you are one step away from solving a quadratic equation. For example, ${x}^{2}-3x-2=0$ can be written as $\left(x-1\right)\left(x-2\right)=0$ . This means that both $x-1=0$ and $x-2=0$ , which gives $x=1$ and $x=2$ as the two solutions to the quadratic equation ${x}^{2}-3x-2=0$ .

1. First divide the entire equation by any common factor of the coefficients, so as to obtain an equation of the form $a{x}^{2}+bx+c=0$ where $a$ , $b$ and $c$ have no common factors. For example, $2{x}^{2}+4x+2=0$ can be written as ${x}^{2}+2x+1=0$ by dividing by 2.
2. Write $a{x}^{2}+bx+c$ in terms of its factors $\left(rx+s\right)\left(ux+v\right)$ . This means $\left(rx+s\right)\left(ux+v\right)=0$ .
3. Once writing the equation in the form $\left(rx+s\right)\left(ux+v\right)=0$ , it then follows that the two solutions are $x=-\frac{s}{r}$ or $x=-\frac{u}{v}$ .
4. For each solution substitute the value into the original equation to check whether it is valid

There are two solutions to a quadratic equation, because any one of the values can solve the equation.

Solve for $x$ : $3{x}^{2}+2x-1=0$

1. As we have seen the factors of $3{x}^{2}+2x-1$ are $\left(x+1\right)$ and $\left(3x-1\right)$ .

2. $\left(x+1\right)\left(3x-1\right)=0$
3. We have

$x+1=0$

or

$3x-1=0$

Therefore, $x=-1$ or $x=\frac{1}{3}$ .

4. We substitute the answers back into the original equation and for both answers we find that the equation is true.
5. $3{x}^{2}+2x-1=0$ for $x=-1$ or $x=\frac{1}{3}$ .

Sometimes an equation might not look like a quadratic at first glance but turns into one with a simple operation or two. Remember that you have to do the same operation on both sides of the equation for it to remain true.

You might need to do one (or a combination) of:

• For example,
$\begin{array}{ccc}\hfill ax+b& =& \frac{c}{x}\hfill \\ \hfill x\left(ax+b\right)& =& x\left(\frac{c}{x}\right)\hfill \\ \hfill a{x}^{2}+bx& =& c\hfill \end{array}$
• This is raising both sides to the power of $-1$ . For example,
$\begin{array}{ccc}\hfill \frac{1}{a{x}^{2}+bx}& =& c\hfill \\ \hfill {\left(\frac{1}{a{x}^{2}+bx}\right)}^{-1}& =& {\left(c\right)}^{-1}\hfill \\ \hfill \frac{a{x}^{2}+bx}{1}& =& \frac{1}{c}\hfill \\ \hfill a{x}^{2}+bx& =& \frac{1}{c}\hfill \end{array}$
• This is raising both sides to the power of 2. For example,
$\begin{array}{ccc}\hfill \sqrt{a{x}^{2}+bx}& =& c\hfill \\ \hfill {\left(\sqrt{a{x}^{2}+bx}\right)}^{2}& =& {c}^{2}\hfill \\ \hfill a{x}^{2}+bx& =& {c}^{2}\hfill \end{array}$

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