How do you solve #sqrt(2x+20) + 2 = x# and find any extraneous solutions?

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Answer 1 · 2017-02-17T04:09:18

#[8}#.

Always start by isolating the radical.

#sqrt(2x+ 20) = x - 2#

Square both sides.

#(sqrt(2x+ 20))^2 = (x - 2)^2#

#2x + 20 = x^2 - 4x + 4#

#0 = x^2 - 6x - 16#

#0 = (x - 8)(x + 2)#

#x = 8 and -2#

The last step to such problems is always to check your solutions within the original equation.

Check: #x= 8#

#sqrt(2(8) +20) + 2 =^? 8#

#sqrt(36) + 2 =8 color(green)(√)#

Check: #x = -2#

#sqrt(2(-2) + 20) + 2 = ^?-2#

#sqrt(16) + 2 != -2#

Therefore, the solution set is #{8}#.

Hopefully this helps!