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Since the first exponent, 1/2, is twice the second exponent, 1/4, then this equation is what is known as "quadratic form". These types of equations can be solved using techniques that are used to solve regular quadratic equations.
Until you have had some practice with these equations they can be difficult. It can help to use a temporary variable. Set it to the variable and exponent in the middle:
Let
Then
Substituting these into your equation we get:
This now actually looks like a quadratic equation. It will factor:
(q+1)(q-2) = 0
From the Zero Product Property:
q+1 = 0 or q-2 = 0
Solving these we get:
q = -1 or q = 2
We have solved for our made-up variable q. But we really want solutions for x. At this point we substitute back in for the q: or
And last of all we solve for x.
The first equation says that a 4th root is equal to a negative number. 4th roots (without a minus in front) must be positive so we will not get any (real) solutions from the first equation.
To solve the second equation we just raise each side to the 4th power:
which simplifies to:
x = 16
Since we raised each side of an equation by an even power we should check. Use the original equation:
Checking x = 16:
Since 1/2 as an exponent means square root and an exponent of 1/4 means 4th root and since we know both the square and 4th roots of 16 this simplifies to:
4 - 2 - 2 = 0
which further simplifies to:
0 = 0 Check!
P.S. Once you get some practice with these you will no longer need the temporary variable. You will start seeing how to go directly from
to
to or
etc.
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