SOLUTION: Could you please help me with this equation? The solution manual says the answer is x=1 but I am getting x = 1 and x= - 4. Could you please help me understand why - 4 is not a so

Algebra ->  Radicals -> SOLUTION: Could you please help me with this equation? The solution manual says the answer is x=1 but I am getting x = 1 and x= - 4. Could you please help me understand why - 4 is not a so      Log On


   

Question 1196335: Could you please help me with this equation? The solution manual says the answer is x=1 but I am getting x = 1 and x= - 4.
Could you please help me understand why - 4 is not a solution? I checked it and it doesn't seem to be an extraneous solution. Would be a huge help if you could please include an explanation for the steps.
Would greatly appreciate your help.
IMG-4646
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
You want to know why -4 is not a solution.

Did you try to test this value?
sqrt%28%28-4%29%28-4%29%2B5%28-4%29%29=sqrt%282%28-4%29%2B4%29
Work with that and see what happens.

Answer by ikleyn(52777) About Me  (Show Source):
You can put this solution on YOUR website!
.

It is  BECAUSE  if you substitute  x= -4  into left side expression,

you will get the number under square root   (-4)^2 +5*(-4) = 16 - 20 = -4.

But negative numbers are not allowed under square root - - - THEREFORE,  we  REJECT   x= - 4.

The same is with the right side expression:  it does not allow  x= -4.

So,  x= -4  is an extraneous solution.


Hope,  everything is clear now.