How do you solve $\sqrt{5 - \frac{x}{4}} = \sqrt{x + 8}$ $\sqrt { 5- \frac { x } { 4} } = \sqrt { x + 8}$ ?

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How do you solve $\sqrt{5 - \frac{x}{4}} = \sqrt{x + 8}$ $\sqrt { 5- \frac { x } { 4} } = \sqrt { x + 8}$ ?

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Answer 1 · 2017-05-08T15:31:48

$x = - \frac{12}{5}$ $x=-12/5$

Explanation:

$\sqrt{5 - \frac{x}{4}} = \sqrt{x + 8}$ $sqrt (5-x/4)=sqrt (x+8)$
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Squaring both sides:
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${(\sqrt{5 - \frac{x}{4}})}^{2} = {(\sqrt{x + 8})}^{2}$ $(sqrt (5-x/4))^2=(sqrt (x+8))^2$
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$\Rightarrow 5 - \frac{x}{4} = x + 8$ $rArr5-x/4=x+8$
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$\Rightarrow - \frac{x}{4} - x = - 5 + 8$ $rArr-x/4-x=-5+8$
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$\Rightarrow - \frac{x}{4} - \frac{4 x}{4} = + 3$ $rArr-x/4-(4x)/4=+3$
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$\Rightarrow - \frac{5 x}{4} = 3$ $rArr-(5x)/4=3$
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$\Rightarrow - 5 x = 12$ $rArr-5x=12$
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$\Rightarrow x = - \frac{12}{5}$ $rArrx=-12/5$

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How do you solve $\sqrt{5 - \frac{x}{4}} = \sqrt{x + 8}$ $\sqrt { 5- \frac { x } { 4} } = \sqrt { x + 8}$ ?

1 Answer

Explanation:

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How do you solve $\sqrt{5 - \frac{x}{4}} = \sqrt{x + 8}$ $\sqrt { 5- \frac { x } { 4} } = \sqrt { x + 8}$ ?