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Answer 1 · 2018-01-31T23:32:37

#x=9#

So we have: #x-1=sqrt (6x+10)#

Let's try to remove the square root sign.

We square both sides.

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

#x^2-2x+1=6x+10# We see that we can form a quadratic equation.

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

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

#-1=x=9#

Before we declare this as our final answer, we test each answer out into our original equation.

#-1-1=sqrt(6*-1+10)#

=>#-2=sqrt(4)#

=>#-2!=2# This tells us that #x=-1# is an extraneous solution.

We check #x=9#

#9-1=sqrt(6*9+10)#

=>#8=sqrt(64)#

=>#8=8# This tells us that #x=9# is a valid solution.

The reason why we have an extraneous solution is because when we square an equation, we have this situation:

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