# How do you solve the rational equation (x^2-7)/(x-4)=0?

Jan 11, 2016

 x = ±sqrt7

#### Explanation:

for this rational function to equate to zero means that the numerator must be zero as the denominator ≠ 0

 rArr x^2 - 7 = 0 rArr x^2 = 7 rArr x = ±sqrt7

Jan 11, 2016

$x = \pm \sqrt{7}$

Strong guidance given on manipulation!

#### Explanation:

Given: $\textcolor{b r o w n}{\textcolor{w h i t e}{\ldots .} \frac{{x}^{2} - 7}{x - 4} = 0}$

Multiply both sides by $\textcolor{b l u e}{\left(x - 4\right)}$

color(brown)((x^2-7)/(x-4)color(blue)(xx(x-4))=0xxcolor(blue)((x-4))

$\left({x}^{2} - 7\right) \times \frac{x - 4}{x - 4} = 0$

But $\frac{x - 4}{x - 4} \text{ is another way of writing } 1$ giving:

$\left({x}^{2} - 7\right) \times 1 = 0$
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Add $\textcolor{b l u e}{7}$ to both sides

color(brown)( x^2-7 color(blue)(+7) =0color(blue)(+7)

${x}^{2} = 7$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

so $\textcolor{w h i t e}{\ldots} x = \sqrt{{x}^{2}} = \pm \sqrt{7}$

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The square root has to be $\pm$ because:

${x}^{2} = \left(+ \sqrt{7}\right) \times \left(+ \sqrt{7}\right) \to \text{positive } {x}^{2}$

And

${x}^{2} = \left(- \sqrt{7}\right) \times \left(- \sqrt{7}\right) \to \text{positive } {x}^{2}$