#sqrt(4x+8)=x+3# ?

#sqrt(4x+8)=x+3#

2 Answers
VNVDVI
Feb 26, 2018

#x=-1#

Explanation:

Square both sides:

#sqrt(4x+8)^2=(x+3)^2#

Squaring a square root causes the square root to cancel out, IE, #sqrt(a)^2=a#, so the left side becomes #4x+8.#

#4x+8=(x+3)^2#

#4x+8=(x+3)(x+3)#

Multiplying out the right side yields:

#4x+8=x^2+6x+9#

We want to solve for #x.# Let's isolate every term on one side and have the other side equal #0.#

#0=x^2+6x-4x+9-8#

#x^2+2x+1=0# (We can switch around our sides since we're working with an equality here. It won't change anything.)

Factoring #x^2+2x+1# yields #(x+1)^2#, as #1+1=2# and #1*1=1.#

#(x+1)^2=0#

Solve for #x# by taking the root of both sides:

#sqrt(x+1)^2=sqrt(0)#

#sqrt(a^2)=a#, so #sqrt(x+1)^2=x+1#

#sqrt(0)=0#

#x+1=0#

#x=-1#

So, #x=-1# may be a solution. We say may be because we must plug #x=-1# into the original equation to make sure our square root is not negative, because negative square roots return non-real answers:

#sqrt(4(-1)+8)=-1+3#

#sqrt(4)=-1+3#

#2=2#

Our root is not negative, so, #x=-1# is the answer.

Jim G.
Feb 26, 2018

#x=-1#

Explanation:

#"square both sides to 'undo' the radical"#

#(sqrt(4x+8))^2=(x+3)^2#

#rArr4x+8=x^2+6x+9#

#"rearrange into "color(blue)"standard form"#

#rArrx^2+2x+1=0#

#rArr(x+1)^2=0#

#rArrx=-1#

#color(blue)"As a check"#

Substitute this value into the original equation and if both sides are equal then it is the solution.

#"left "=sqrt(-4+8)=sqrt4=2#

#"right "=-1+3=2#

#rArrx=-1" is the solution"#

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