SOLUTION: The solution of sqrt(x+4)= x-2 ==> x=0 and x=5 x=5 is true since by substituting in the original equation yield 3=3 However,for x=0 we have sqrt(4)=-2 which give an incorre

Algebra ->  Radicals -> SOLUTION: The solution of sqrt(x+4)= x-2 ==> x=0 and x=5 x=5 is true since by substituting in the original equation yield 3=3 However,for x=0 we have sqrt(4)=-2 which give an incorre      Log On


   



Question 719384: The solution of
sqrt(x+4)= x-2 ==> x=0 and x=5
x=5 is true since by substituting in the original equation yield 3=3
However,for x=0 we have sqrt(4)=-2 which give an incorrect solution... If we ignore the negative value of squrt(4).
My question is.. Why the negative value is ignored?

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Any positive number has two square roots, a positive one and a negative one. If "a" is a positive number then
  • sqrt%28a%29 is a specific reference to just the positive square root.
  • -sqrt%28a%29 is a specific reference to just the negative square root.
  • +sqrt%28a%29 (like you see in the quadratic formula) is a way to reference both of the square roots.

So the sqrt%28x%2B4%29 is given to you in this problem is a reference to just the positive square root. We are not ignoring the negative square root of x+4 because the problem has no reference to the negative square root of 4: -sqrt%28x%2B4%29.

x = 0 is an incorrect solution not because we're ignoring a square root but because the equation is not true when x = 0

P.S. There is an important difference between square roots that are given in a problem and square roots you add while trying to solve a problem. For example, if you're trying to solve:
y%5E2+=+7
You could find the square root of each side. Since you are adding the square roots, you are responsible for not ignoring the negative square root. Both the positive and negative square roots of 7 will be solutions and we cannot forget that. So you can't just say:
y = sqrt%287%29 because that square root is a reference to just the positive root. We have to say
y = +sqrt%287%29 to refer to both square roots at once, or say:
y = sqrt%287%29 or y = -sqrt%287%29
to refer to them both separately.