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The key to making this problem solvable is to recognize that the exponent on is twice as bug as the exponent on . This makes your equation in "quadratic form" for . So we can use techniques for solving quadratic equations to solve for !
To make this process clearer, I am going to use a temporary variable. Let q = . Then . Substituting these into the equation we get:
This is clearly a quadratic equation. With the square root in there, factoring will not be easy. So we'll use the Quadratic formula:
Simplifying:
In long form: or
This is a solution for q. But we want a solution for x. So we substitute back in for q: or
Looking at the second equation above we see that the fraction is negative. But a negative numbers cannot be equal to something squared, like ! So unless you are looking for complex solutions, too, there are no solutions to the second equation. We are left with just the first equation:
To solve for x we just find the square root of each side:
(Note: The zero is unnecessary mathematically. But Algebra.com's software will not let me use the "plus or minus" symbol without a number in front of it.)
In long form: or
Eventually, after doing several of these "quadratic form"-type equations, you will be able to do these problems without a temporary variable. IOW, you will be able to straight from
to
(