SOLUTION: solve the radical equation sqrt 2x+8=x+4. I have no idea where to start

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Question 396176: solve the radical equation sqrt 2x+8=x+4. I have no idea where to start
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Vocabulary note - Radicand: the expression within a radical

Here's a step-by-step procedure for solving equations where the variable is in the radicand of a square root (like yours):
  1. Isolate a square root that has a variable in its radicand. ("Isolate" means get it by itself on one side of the equation.)
  2. Square both sides of the equation.
  3. If there is still a square root with a variable in its radicand, then repeat steps 1 and 2 until they are all gone.
  4. At this point there should be no square roots with a variable in its radicand. Use appropriate techniques to solve whatever type of equation you mow have.
  5. Check your answers. This is required. Whenever you square both sides of an equation, which has been done at least once so far, then extraneous solutions may appear. Extraneous solutions are solutions which fit the squared equation but do not fit the original equation! Extraneous solutions can appear even if you have no errors! So you must check your solutions in the original equation and if you find any extraneous solutions you reject them.

Let's see how this works on your equation:
sqrt%282x%2B8%29=x%2B4
1) Isolate a square root. Your equation only has one square root and it is already isolated.
2) Square both sides:
%28sqrt%282x%2B8%29%29%5E2=%28x%2B4%29%5E2
The left side is easy to square. On the right side we must use FOIL or the %28a%2Bb%29%5E2+=+a%5E2+%2B+2ab+%2B+b%5E2 pattern to square it properly. I prefer the pattern:
2x+%2B+8+=+%28x%29%5E2+%2B+2%28x%29%284%29+%2B+%284%29%5E2
which simplifies to:
2x+%2B+8+=+x%5E2+%2B+8x+%2B+16
3) Repeat steps 1 and 2 if there are still any square roots with variable in them.
We have no square roots left.
4) Solve the resulting equation.
This is a quadratic equation (because of the x%5E2) so we want one side to be zero. Subtracting 2x and 8 from each side:
0+=+x%5E2+%2B+6x+%2B+8
Then we factor (or use the Quadratic Formula). This factors easily:
0 = (x+2)(x+4)
Then we use the Zero Product Property which tells us that this (or any) product can be zero only if one (or more) of the factor is zero. So:
x+2 = 0 or x+4 = 0
Solving these we get:
x = -2 or x = -4
5) Check your solutions (using the original equation):
sqrt%282x%2B8%29=x%2B4
Checking x = -2:
sqrt%282%28-2%29%2B8%29=%28-2%29%2B4
which simplifies as follows:
sqrt%28-4%2B8%29=-2%2B4
sqrt%284%29=2
2=2 Check!
Checking x = -4:
sqrt%282%28-4%29%2B8%29=%28-4%29%2B4
which simplifies as follows:
sqrt%28-8%2B8%29=-4%2B4
sqrt%280%29=0
0=0 Check!
Both solutions checked. (But we did not know they would until we actually checked. One or even both of them could have been extraneous solutions!)