To solve equations with a variable expression raised to a rational exponent, we first want to make sure the variable expression is isolated. Then we raise each side of the equation to the reciprocal of the rational exponent.
xnm=k⇔(xnm)mn=kmn
Remember, if m is even, then (xnm)mn=∣x∣.
In this case m=1, so we do not need to worry about this.
Finding the Solutions
We will first isolate the variable expression raised to a rational exponent, and then raise each side of the equation to the power of 2.
Using the Quadratic Formula, we found that the solutions of the given equation are x=25±3. Therefore, the solutions are x1=1 and x2=4.
Let's check them to see if we have any extraneous solutions.
Checking the Solutions
We will check x1=1 and x2=4 one at a time. Let's start by substituting x=1 into the original equation.
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