Factorisation

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Introduction

In grade 10, the basics of solving linear equations, quadratic equations, exponential equations and linear inequalities were studied. This chapter extends on that work. We look at different methods of solving quadratic equations.

Solution by factorisation

How to solve quadratic equations by factorisation was discussed in Grade 10. Here is an example to remind you of what is involved.

Solve the equation $2{x}^{2}-5x-12=0$ .

1. This equation has no common factors.

2. The equation is in the required form, with $a=2$ , $b=-5$ and $c=-12$ .

3. $2{x}^{2}-5x-12$ has factors of the form:

$\left(2x+s\right)\left(x+v\right)$

with $s$ and $v$ constants to be determined. This multiplies out to

$2{x}^{2}+\left(s+2v\right)x+sv$

We see that $sv=-12$ and $s+2v=-5$ . This is a set of simultaneous equations in $s$ and $v$ , but it is easy to solve numerically. All the options for $s$ and $v$ are considered below.

 $s$ $v$ $s+2v$ 2 -6 -10 -2 6 10 3 -4 -5 -3 4 5 4 -3 -2 -4 3 2 6 -2 2 -6 2 -2

We see that the combination $s=3$ and $v=-4$ gives $s+2v=-5$ .

4. $\left(2x+3\right)\left(x-4\right)=0$
5. If two brackets are multiplied together and give 0, then one of the brackets must be 0, therefore

$2x+3=0$

or

$x-4=0$

Therefore, $x=-\frac{3}{2}$ or $x=4$

6. The solutions to $2{x}^{2}-5x-12=0$ are $x=-\frac{3}{2}$ or $x=4$ .

It is important to remember that a quadratic equation has to be in the form $a{x}^{2}+bx+c=0$ before one can solve it using the following methods.

Solve for $a$ : $a\left(a-3\right)=10$

1. Remove the brackets and move all terms to one side.

${a}^{2}-3a-10=0$
2. $\left(a+2\right)\left(a-5\right)=0$
3. $a+2=0$

or

$a-5=0$

Solve the two linear equations and check the solutions in the original equation.

4. Therefore, $a=-2$ or $a=5$

Solve for $b$ : $\frac{3b}{b+2}+1=\frac{4}{b+1}$

1. $\frac{3b\left(b+1\right)+\left(b+2\right)\left(b+1\right)}{\left(b+2\right)\left(b+1\right)}=\frac{4\left(b+2\right)}{\left(b+2\right)\left(b+1\right)}$
2. The denominators are the same, therefore the numerators must be the same.

However, $b\ne -2$ and $b\ne -1$

3. $\begin{array}{ccc}\hfill 3{b}^{2}+3b+{b}^{2}+3b+2& =& 4b+8\hfill \\ \hfill 4{b}^{2}+2b-6& =& 0\hfill \\ \hfill 2{b}^{2}+b-3& =& 0\hfill \end{array}$
4. $\begin{array}{ccc}\hfill \left(2b+3\right)\left(b-1\right)& =& 0\hfill \\ \hfill 2b+3=0& or& b-1=0\hfill \\ \hfill b=\frac{-3}{2}& or& b=1\hfill \end{array}$
5. Both solutions are valid

Therefore, $b=\frac{-3}{2}$ or $b=1$

Solution by factorisation

Solve the following quadratic equations by factorisation. Some answers may be left in surd form.

1. $2{y}^{2}-61=101$
2. $2{y}^{2}-10=0$
3. ${y}^{2}-4=10$
4. $2{y}^{2}-8=28$
5. $7{y}^{2}=28$
6. ${y}^{2}+28=100$
7. $7{y}^{2}+14y=0$
8. $12{y}^{2}+24y+12=0$
9. $16{y}^{2}-400=0$
10. ${y}^{2}-5y+6=0$
11. ${y}^{2}+5y-36=0$
12. ${y}^{2}+2y=8$
13. $-{y}^{2}-11y-24=0$
14. $13y-42={y}^{2}$
15. ${y}^{2}+9y+14=0$
16. ${y}^{2}-5ky+4{k}^{2}=0$
17. $y\left(2y+1\right)=15$
18. $\frac{5y}{y-2}+\frac{3}{y}+2=\frac{-6}{{y}^{2}-2y}$
19. $\frac{y-2}{y+1}=\frac{2y+1}{y-7}$

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