# 2.2 Solving quadratic equations  (Page 2/2)

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You can combine these in many ways and so the best way to develop your intuition for the best thing to do is practice problems. A combined set of operations could be, for example,

$\begin{array}{cccc}\hfill \frac{1}{\sqrt{a{x}^{2}+bx}}& =& c\hfill & \\ \hfill {\left(\frac{1}{a{x}^{2}+bx}\right)}^{-1}& =& {\left(c\right)}^{-1}\hfill & \left(\mathrm{invert both sides}\right)\hfill \\ \hfill \frac{\sqrt{a{x}^{2}+bx}}{1}& =& \frac{1}{c}\hfill & \\ \hfill \sqrt{a{x}^{2}+bx}& =& \frac{1}{c}\hfill & \\ \hfill {\left(\sqrt{a{x}^{2}+bx}\right)}^{2}& =& {\left(\frac{1}{c}\right)}^{2}\hfill & \left(\mathrm{square both sides}\right)\hfill \\ \hfill a{x}^{2}+bx& =& \frac{1}{{c}^{2}}\hfill & \end{array}$

Solve for $x$ : $\sqrt{x+2}=x$

1. Both sides of the equation should be squared to remove the square root sign.

$x+2={x}^{2}$
2. $\begin{array}{cccc}\hfill x+2& =& {x}^{2}\hfill & \left(\mathrm{subtract}\phantom{\rule{2pt}{0ex}}{\mathrm{x}}^{2}\phantom{\rule{2pt}{0ex}}\mathrm{from both sides}\right)\hfill \\ \hfill x+2-{x}^{2}& =& 0\hfill & \left(\mathrm{divide both sides by}\phantom{\rule{2pt}{0ex}}-1\right)\hfill \\ \hfill -x-2+{x}^{2}& =& 0\hfill & \\ \hfill {x}^{2}-x+2& =& 0\hfill & \end{array}$
3. ${x}^{2}-x+2$

The factors of ${x}^{2}-x+2$ are $\left(x-2\right)\left(x+1\right)$ .

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

$x+1=0$

or

$x-2=0$

Therefore, $x=-1$ or $x=2$ .

6. Substitute $x=-1$ into the original equation $\sqrt{x+2}=x$ :

$\begin{array}{ccc}\hfill \mathrm{LHS}& =& \sqrt{\left(-1\right)+2}\hfill \\ & =& \sqrt{1}\hfill \\ & =& 1\hfill \\ \hfill \mathrm{but}\\ \hfill \mathrm{RHS}& =& \left(-1\right)\hfill \end{array}$

Therefore LHS RHS. The sides of an equation must always balance, a potential solution that does not balance the equation is not valid. In this case the equation does not balance.

Therefore .

Now substitute $x=2$ into original equation $\sqrt{x+2}=x$ :

$\begin{array}{ccc}\hfill \mathrm{LHS}& =& \sqrt{2+2}\hfill \\ & =& \sqrt{4}\hfill \\ & =& 2\hfill \\ \hfill \mathrm{and}\\ \hfill \mathrm{RHS}& =& 2\hfill \end{array}$

Therefore LHS = RHS

Therefore $x=2$ is the only valid solution

7. $\sqrt{x+2}=x$ for $x=2$ only.

Solve the equation: ${x}^{2}+3x-4=0$ .

1. The equation is in the required form, with $a=1$ .

2. You need the factors of 1 and 4 so that the middle term is $+3$ So the factors are:

$\left(x-1\right)\left(x+4\right)$

3. ${x}^{2}+3x-4=\left(x-1\right)\left(x+4\right)=0$

Therefore $x=1$ or $x=-4$ .

4. ${1}^{2}+3\left(1\right)-4=0$
${\left(-4\right)}^{2}+3\left(-4\right)-4=0$

Both solutions are valid.

5. Therefore the solutions are $x=1$ or $x=-4$ .

Find the roots of the quadratic equation $0=-2{x}^{2}+4x-2$ .

1. There is a common factor: -2. Therefore, divide both sides of the equation by -2.

$\begin{array}{ccc}\hfill -2{x}^{2}+4x-2& =& 0\hfill \\ \hfill {x}^{2}-2x+1& =& 0\hfill \end{array}$
2. The middle term is negative. Therefore, the factors are $\left(x-1\right)\left(x-1\right)$

If we multiply out $\left(x-1\right)\left(x-1\right)$ , we get ${x}^{2}-2x+1$ .

3. ${x}^{2}-2x+1=\left(x-1\right)\left(x-1\right)=0$

In this case, the quadratic is a perfect square, so there is only one solution for $x$ : $x=1$ .

4. $-2{\left(1\right)}^{2}+4\left(1\right)-2=0$ .

5. The root of $0=-2{x}^{2}+4x-2$ is $x=1$ .

## Solving quadratic equations

1. Solve for $x$ : $\left(3x+2\right)\left(3x-4\right)=0$
2. Solve for $x$ : $\left(5x-9\right)\left(x+6\right)=0$
3. Solve for $x$ : $\left(2x+3\right)\left(2x-3\right)=0$
4. Solve for $x$ : $\left(2x+1\right)\left(2x-9\right)=0$
5. Solve for $x$ : $\left(2x-3\right)\left(2x-3\right)=0$
6. Solve for $x$ : $20x+25{x}^{2}=0$
7. Solve for $x$ : $4{x}^{2}-17x-77=0$
8. Solve for $x$ : $2{x}^{2}-5x-12=0$
9. Solve for $x$ : $-75{x}^{2}+290x-240=0$
10. Solve for $x$ : $2x=\frac{1}{3}{x}^{2}-3x+14\frac{2}{3}$
11. Solve for $x$ : ${x}^{2}-4x=-4$
12. Solve for $x$ : $-{x}^{2}+4x-6=4{x}^{2}-5x+3$
13. Solve for $x$ : ${x}^{2}=3x$
14. Solve for $x$ : $3{x}^{2}+10x-25=0$
15. Solve for $x$ : ${x}^{2}-x+3$
16. Solve for $x$ : ${x}^{2}-4x+4=0$
17. Solve for $x$ : ${x}^{2}-6x=7$
18. Solve for $x$ : $14{x}^{2}+5x=6$
19. Solve for $x$ : $2{x}^{2}-2x=12$
20. Solve for $x$ : $3{x}^{2}+2y-6={x}^{2}-x+2$

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Source:  OpenStax, Maths grade 10 rought draft. OpenStax CNX. Sep 29, 2011 Download for free at http://cnx.org/content/col11363/1.1
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