# Question #01a12

May 17, 2016

Supporting detail to solution provided by Brian

$x = \pm 2$

#### Explanation:

Given:$\text{ } \left({x}^{2} + 2\right) = 6$

In general the purpose of brackets is that of grouping values for clarity or for some form of operation to be carried out on the whole group.

In the given equation no such operation is applied nor is there any need to make anything more clear. So they are redundant.

Thus you can dispense with the brackets giving:

${x}^{2} + 2 = 6$

Subtract 2 from both sides

${x}^{2} + 2 - 2 = 6 - 2$

But +2-2=0 giving

${x}^{2} = 4$

square root both sides

$\sqrt{{x}^{2}} = \sqrt{4}$

$x = 2$
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Actually $x = \pm 2$

as $\left(- 2\right) \times \left(- 2\right) = + 4 \text{ and } \left(+ 2\right) \times \left(+ 2\right) = + 4$