Question 760427
To solve we need to find a way to address the absolute values. We can use the definition:


{{{
abs(x^2-8)=abs(9-x)-1
}}}


that means there are four possibilities:


1)

{{{
x^2-8=9-x-1
}}}


which leads to

{{{
x = (-1 +- sqrt(65))/2
 }}}

2)
{{{
x^2-8=x-9-1
}}}

which yields complex roots

3)
{{{
x^2-8=9-x+1
}}}

Which yields extraneous roots

4)
{{{
x^2-8=x-9+1
}}}

which yields 
x=0, x=1


:)