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Since the exponent of is exactly twice the exponent of this equation is in what is called quadratic form. Equations in quadratic form can be solved in much the same way as "regular" quadratic equations.
Until you get used to these quadratic form equations, it can be helpful to use a temporary variable:
Let
Then
Substituting these into the equation we get:
This is obviously a quadratic equation. And factoring is fairly easy:
(q-9)(q-1) = 0
From the Zero Product Property we know that one of the factors must be zero. So:
q-9 = 0 or q-1 = 0
Solving these we get:
q = 9 or q = 1
Of course we are not interested in what "q" is equal to. We are interested in what "y" is equal to. So at this point we substitute back in for the q's. (Remember, q was only a temporary variable.) Replacing the q's with (which is what we defined q to be equal to) we get: or
To solve for y we need to do a little more work. You could just find the square root of each side. But you have to remember that there are positive and negative square roots of 9 and 1. Since this is easy to forget I like to solve these equations another way. First we'll make one side zero: or
Then factor:
(y+3)(y-3) = 0 or (y+1)(y-1) = 0
Then use the Zero Product Property:
y+3 = 0 or y-3 = 0 or y+1 = 0 or y-1 = 0
Solving these we get:
y = -3 or y = 3 or y = -1 or y = 1
Although this method is a little more work than just finding square roots, it does handle the positive and negative square roots automatically.
P.S. Once you get used to these quadratic form equations, you will no longer need a temporary variable. You will see how to go directly from
to
to
to
y+3 = 0 or y-3 = 0 or y+1 = 0 or y-1 = 0
to
y = -3 or y = 3 or y = -1 or y = 1
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