SOLUTION: I was needing to check another answer, the problem is, sqrt(x+10)+2=x my answer was x=-6 andx=-1 but -1 is not a solution

Question 173483: I was needing to check another answer, the problem is, sqrt(x+10)+2=x my answer was x=-6 andx=-1 but -1 is not a solution
Found 3 solutions by Alan3354, Mathtut, solver91311:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
I was needing to check another answer, the problem is, sqrt(x+10)+2=x my answer was x=-6 andx=-1 but -1 is not a solution
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sqrt(x+10)+2=x
sqrt(-1 + 10) +2 =? -1
sqrt(9) + 2 =? -1
-3 + 2 = -1
Looks ok to me.

Answer by Mathtut(3670)   (Show Source): You can put this solution on YOUR website!

:

:
squaring both sides
:

:

:
it works in the quad equation but the only way it works in the original is if the original left side had a negative in front of the radical
:
When solving radical equations, extra solutions may come up when you raise both sides to an even power. These extra solutions are called extraneous solutions. If a value is an extraneous solution, it is not a solution to the original problem.
In radical equations, you check for extraneous solutions by plugging in the values you found back into the original problem. If the left side does not equal the right side, then you have an extraneous solution.
:therefore -1 is not a solution to this equation
:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=49 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 6, -1. Here's your graph:

Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

I think you probably had the process correct, but you made a sign error.

Add -2 to both sides:

Square both sides:

Collect terms and put in standard form:

and so:

=> or

=>

is a valid root because but

is NOT a valid root because

The invalid, or extraneous root was introduced by the act of squaring the variable in the process of solving the equation. You must ALWAYS check your answers against the original equation any time you square a variable in the process of solving.
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