How do you solve $\sqrt { 2x + 48} = x$ ?

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Answer 1 · 2016-12-16T02:22:31

${8}$ .

Square both sides:

$(sqrt(2x +48))^2 = x^2$

$2x + 48 = x^2$

$0 = x^2 - 2x - 48$

$0 = (x - 8)(x + 6)$

$x= 8 and -6$

Check, as extraneous solutions may have been introduced in the solving process.

$sqrt(2(8) + 48) =^? 8$

$sqrt(64) = 8" "color(green)(√)$

AND

$sqrt(2(-6) + 48) =^? -6$

$sqrt(36) != -6" "color(red)(xx)$

Hopefully this helps!

How do you solve \sqrt { 2x + 48} = x\sqrt { 2x + 48} = x?