You can put this solution on YOUR website!
The key to the easiest solution to this equation is to recognize that there are two exponents and that one of them is twice the other. (1/2 is twice 1/4) This makes the equation and equation of "quadratic form".
Equations of quadratic form can be solved using the same techniques as those for "regular" quadratic equations. So we start by making one side zero. (And just like quadratic equations, it will be easier if we keep the coefficient in front of larger exponent positive.) So I'm going to subtract the entire right side from each side:
Until you get comfortable with these quadratic form equations, using a temporary variable can make it easier. Set the temporary variable equal to the expression with the lower exponent. So:
Substituting these into the equation we get:
This is clearly a quadratic equation. We now factor (or use the Quadratic Formula). This factors fairly easily:
(q-9)(q-2) = 0
Fron the Zero Product Property we know that one of these factors must be zero. So:
q-9 = 0 or q-2 = 0
Solving these we get:
q = 9 or q = 2
Of course we are not interested in what q is. We want to know what x is. So we substitute back in for the temporary variables:
To solve these we raise each side to the 4th power:
which simplify to:
x-2 = 6561 or x-2 = 16
Adding 2 we get:
x = 6563 or x = 18
These are the solutions to your equation.
After a few of these you will not need the temporary variable. You will see how to go from