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!