Question 355081
{{{(x^2+9)^(3/2)=125}}}
First of all we need to get rid of that exponent of 3/2. How do you make an exponent "disappear"? When do exponents become "invisible"? The answer to both questions is: when the exponent becomes a 1!<br>
So we need to think about how we can use proper Math to change an exponent of 3/2 into a 1. We should know that whenever you multiply reciprocals you get a 1. And you should know that there is a rule for exponents, {{{(a^p)^q = a^((p*q))}}}, which tels you when it is correct to multiply exponents. Putting these together, we find that raising the 3/2 power to the 2/3 (2/3 is the resiprocal of 3/2) power then the exponent will become a 1 and "disappear"! So we will raise both sides of this equation to the 2/3 power:
{{{((x^2+9)^(3/2))^(2/3)=(125)^(2/3)}}}
Simplifying on the left is easy since both exponents combine to form a 1 which "disappears". On the right side we want to raise 125 to the 2/3 power. In the  exponent of 2/3 the 2 means we will square something and the 1/3 means we will find a cube root of something. And we can do these in any order. Since {{{125 = 5^3}}}, I'm going to find the cube root first:
{{{((x^2+9)^(3/2))*(2/3)=(125)^(2*(1/3))}}}
{{{(x^2+9)^1=(125^(1/3))^2}}}
{{{x^2+9=5^2}}}
{{{x^2+9=25}}}
Now we have a quadratic equation to solve. Subtract 25 from each side:
{{{x^2 - 16 = 0}}}
Factor the difference of squares:
{{{(x + 4)(x - 4) = 0}}}
From the Zero Product Property we know that this (or any) product can be zero <i>only</i> if one of the factors is zero. So:
x + 4 = 0 or x - 4 = 0
Solving these we get:
x = -4 or x = 4