# How do you solve x^2-5=0 graphically and algebraically?

Mar 20, 2016

$x = \pm \sqrt{5}$ exactly, or approximately $\pm 2.236$ to 3 decimal places

#### Explanation:

Given:$\text{ } {x}^{2} = 5 = 0$

To solve mathematically we need to have just 1 $x$ and for it to be on one side of the equals sign and everything else on the other side.

To get the x term on its own.

Add $\textcolor{b l u e}{5}$ to both sides of the equation

" "color(brown)(x^2-5color(blue)(+5)=0color(blue)(+5)

${x}^{2} + 0 = 5$

${x}^{2} = 5$

But ${x}^{2}$ is the same as $x \times x$ and we need to have just 1 not 2 of them.

Take the square root of both sides

$\sqrt{{x}^{2}} = \sqrt{5}$

So $x = \sqrt{5} \text{ }$ which is 2.236 to 3 decimal places

But both $\left(- 2.236\right) \times \left(- 2.236\right) = 5 \text{ }$

and$\text{ } \left(+ 2.236\right) \times \left(+ 2.236\right) = 5$

So $x = \pm \sqrt{5}$ exactly or approximately $\pm 2.236$ to 3 decimal places

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