Noticing that the exponent of is twice as large as the exponent for .
Understanding that any equation with a pattern of exponents like this (one exponent is twice the other) is an equation of "quadratic form".
Equations of quadratic form can be solved, in part, using the same methods as those for quadratic equations.
After you have done a few of these quadratic form equations, they get easier. Until then you might find it helpful to use a temporary variable:
Let
Then
Substituting these into your equation we get:
This is obviously a quadratic equation. We can solve it by factoring (or using the Quadratic Formula). This factors fairly easily:
(4q - 3)(q - 1) = 0
This product can be zero only if one of the factors is zero. So:
4q-3 = 0 or q-1 = 0
Solving these we get:
q = 3/4 or q = 1
We now have solutions for q. But what we want are solutions for x. So we substitute back in for q: or
To solve for x we raise both sides of each equation to the 3/4 power: or
On the left sides, the rule for exponents when raising a power to a power is to multiply the exponents. And multiplying 4/3 times 3/4 results in a 1! (This is why we used 3/4, the reciprocal of 4/3. Multiplying reciprocals always results in a 1 and we want an exponent of 1 on x!) or
or or
The right side of the second equation is easy since 1 to any power is 1. The right side of the first equation is not so easy. We will start by rewriting it in radical form: or
Cubing 3/4 we get: or
We want the denominator of the fraction to be a power of 4. so we just need one more factor of 4: or or or or
These are the solutions to your equation.
As I mentioned earlier, these "quadratic form" equations take some getting used to. But eventually you will be able to see how to solve them without the temporary variable. You will see how to go directly from:
to
to or
etc.
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