How do you solve #x-1= sqrt( 6x+10)#?

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Answer 1 · 2016-11-07T14:05:24

#x = 9#.

#(x - 1)^2 = (sqrt(6x + 10))^2#

#x^2 - 2x + 1 = 6x + 10#

#x^2 - 8x - 9 = 0#

#(x - 9)(x + 1) = 0#

#x = 9 and x =-1#

However, since extraneous solutions are a distinct possibility, let's check our solution inside the initial equation.

#9 - 1 =^? sqrt(6 xx 9 + 10)#

#8 = 8" "color(green)(√)#

AND

#-1 - 1=^? sqrt(6 xx -1 + 10)#

#-2 != 2" "color(red)(xx)#

Hence, the only true solution is #x = 9#.

Hopefully this helps!