SOLUTION: Find all the real solutions to the equation: |x^(2) -8|=|9-x|-1. Provide exact solutions (not rounded numbers) please and thank you!

Algebra ->  Absolute-value -> SOLUTION: Find all the real solutions to the equation: |x^(2) -8|=|9-x|-1. Provide exact solutions (not rounded numbers) please and thank you!      Log On


   

Question 760427: Find all the real solutions to the equation: |x^(2) -8|=|9-x|-1. Provide exact solutions (not rounded numbers) please and thank you!
Found 2 solutions by josgarithmetic, tommyt3rd:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2-8 could be either {positive or 0} or {negative}.
9-x could be either {positive or 0} or {negative}.
Those conditions give you four different possible equations to solve.

I will show you just one of the possible equations and best if you setup the other three and solve. ... then consider checking each result in the original equation.

IF both expressions are negative, then you use
%28-x%5E2%2B8%29=%28x-9%29%2B1
and solve this quadratic equation for x.

Now ...
...
!

Answer by tommyt3rd(5050) About Me  (Show Source):
You can put this solution on YOUR website!
To solve we need to find a way to address the absolute values. We can use the definition:

%0D%0Aabs%28x%5E2-8%29=abs%289-x%29-1%0D%0A

that means there are four possibilities:

1)
%0D%0Ax%5E2-8=9-x-1%0D%0A

which leads to
%0D%0Ax+=+%28-1+%2B-+sqrt%2865%29%29%2F2%0D%0A+
2)
%0D%0Ax%5E2-8=x-9-1%0D%0A
which yields complex roots
3)
%0D%0Ax%5E2-8=9-x%2B1%0D%0A
Which yields extraneous roots
4)
%0D%0Ax%5E2-8=x-9%2B1%0D%0A
which yields
x=0, x=1

:)