How do you solve # (x^2 - 5)/(x + 3) = 0#?

1 Answer
Mar 26, 2016

I found: #x=+-sqrt(5)#

Explanation:

We need to find the value(s) of #x# that satisfy our equation; we also need to avoid #x# values that make our relationship INDETERMINATE!
This is because if you make the denominator zero (choosing #x=-3#) the division by zero is not possible!
So as first consideration we say that:
#x!=-3#
Next we solve our equation:

#x^2-5=0*(x+3)#
moving the denominator to the right of the equal sign;
so:
#x^2-5=0#
#x^2=5#
#x=+-sqrt(5)#