How do you solve #sqrt (4x + 8) - 1 = x - 2#?

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Answer 1 · 2016-01-26T04:11:25

1) leave the square root alone on 1 side:
#sqrt(4x+8) = x-2+1 = x-1#

2) square both sides:
#4x+8 = (x-1)^2#

3) Expand the right side:#(a+b)^2=a^2+2ab+b^2#
#4x-8 = x^2 -2x+1#

4) Move everything to one side and you get:
#x^2 -6x +9 = 0#

5) This. luckily factors out to:
#(x-3)^2=0#
so #x=3#

Answer 2 · 2016-01-31T07:08:42

#x=7,-1#

#sqrt(4x+8)-1=x-2#

#rarrsqrt(4x+8)=x-2+1#

#rarrsqrt(4x+8)=x-1#

Square both sides:

#rarr(sqrt(4x-8))^2=(x-1)^2#

Use the formula #(a+b)^2=a^2+2ab+b^2# for the equation on the R.h.s

#rarr4x-8=x^2-2x+1#

#rarr0=x^2-2x+1-4x8#

#rarr0=x^2-6x-7#

Now this is a quadratic equation:Solve using quadratic formula:

#rarrx=7,-1#