How do you solve ${(2 x)}^{\frac{1}{2}} = x - 4$ $(2x)^(1/2)=x-4$ ?

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Answer 1 · 2017-02-22T16:50:13

${8}$ ${8}$

You're going to want to get rid of the 1/2 exponent. You can do this by squaring both sides.

${({(2 x)}^{\frac{1}{2}})}^{2} = {(x - 4)}^{2}$ $((2x)^(1/2))^2 = (x - 4)^2$

$2 x = (x - 4) (x - 4)$ $2x = (x - 4)(x -4)$

$2 x = x^{2} - 4 x - 4 x + 16$ $2x = x^2 - 4x - 4x + 16$

$0 = x^{2} - 10 x + 16$ $0 = x^2 - 10x + 16$

$0 = (x - 8) (x - 2)$ $0 = (x - 8)(x - 2)$

$x = 8 and 2$ $x = 8 and 2$

This is a rational equation, so extraneous solutions are always a possibility, if not a likelihood. Always check your solutions.

${(2 \cdot 2)}^{\frac{1}{2}} =^{?} 2 - 4$ $(2 * 2)^(1/2) =^? 2 - 4$

$\sqrt{2} \neq - 2 \times$ $sqrt(2) != -2 color(red)(xx)$

${(2 \cdot 8)}^{\frac{1}{2}} =^{?} 8 - 4$ $(2 * 8)^(1/2) =^? 8 - 4$

$\sqrt{16} = 4$ $sqrt(16) = 4$

The only valid solution is $x = 8$ $x = 8$ .

Hopefully this helps!

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