SOLUTION: Any help would be appreciated: (sqrt2x+7)-(x+2)=0 I'm not sure if I wrote it properly, but it's the square root of 2x+7.

Algebra ->  Radicals -> SOLUTION: Any help would be appreciated: (sqrt2x+7)-(x+2)=0 I'm not sure if I wrote it properly, but it's the square root of 2x+7.      Log On


   

Question 423062: Any help would be appreciated:
(sqrt2x+7)-(x+2)=0
I'm not sure if I wrote it properly, but it's the square root of 2x+7.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%282x%2B7%29-%28x%2B2%29+=+0
Here's a procedure for solving square root equations:
  1. Isolate a square root (with a variable in its radicand). (The expression inside a radical is called a radicand.)
  2. Square both sides of the equation (carefully).
  3. If there is still a square root (with a variable in its radicand), then repeat steps 1 and 2.
  4. At this point there should be no square roots (that have a variable in its radciand). Use appropriate techniques to solve the equation.
  5. Check your answer(s)! This is not optional. When you square both sides of an equation, which has happened at least once at step 2, extraneous solutions may be introduced. Extraneous solutions are solutions are solutions that fit the squared equation but do not fit the original equation. Extraneous solutions may happen anytime you square both sides. They do not mean it is a mistake to square both sides! So you must check your answers and make sure they fit the original equation. Any "solutions" that do not fit the original equation are extraneous and must be rejected.
Let's see how this works:
1) Isolate a square root...
There's only one square root. To isolate it we will simplify:
sqrt%282x%2B7%29-x-2+=+0
and then add x and 2 to each side:
sqrt%282x%2B7%29+=+x%2B2

2) Square both sides.
%28sqrt%282x%2B7%29%29%5E2+=+%28x%2B2%29%5E2
Squaring the left side is easy. Squaring the right side, correctly, is done by using FOIL or by using the %28a%2Bb%29%5E2+=+a%5E2%2B2ab%2Bb%5E2 pattern. I prefer using the pattern:
2x%2B7+=+%28x%29%5E2%2B2%28x%29%282%29%2B%282%29%5E2
which simplifies to:
2x%2B7+=+x%5E2%2B4x%2B4

3) If there are still square roots...
There are no square roots.

4) Solve the equation.
This is a quadratic equation. So we want one side of the equation to be zero. Subtracting 2x and 7 from both sides we get:
0+=+x%5E2%2B2x-3
Now we factor (or use the Quadratic Formula). This factors easily:
0 = (x+3)(x-1)
From the Zero Product Property we know that one of these factors must be zero. So:
x+3 = 0 or x-1 = 0
Solving these we get:
x = -2 or x = 1

5) Check your answer(s).
Remember this is not optional. Use the original equation to check:
sqrt%282x%2B7%29-%28x%2B2%29+=+0
Checking x = -3:
sqrt%282%28-3%29%2B7%29-%28%28-3%29%2B2%29+=+0
which simplifies as follows:
sqrt%28-6%2B7%29-%28-1%29+=+0
sqrt%281%29-%28-1%29+=+0
1-%28-1%29+=+0
1%2B1+=+0
2+=+0 Check failed!!
This is an extraneous solution which we reject.

Checking x = 1:
sqrt%282%281%29%2B7%29-%28%281%29%2B2%29+=+0
sqrt%282%2B7%29-%283%29+=+0
sqrt%289%29-%283%29+=+0
3-%283%29+=+0
0+=+0 Check!

So the only solution to your equation is: x = 1