SOLUTION: solve with rational exponents then check solution...???... (x^2-x-4)^3/4-2=6

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Question 345185: solve with rational exponents then check solution...???...

(x^2-x-4)^3/4-2=6

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%5E2-x-4%29%5E%283%2F4%29-2=6
First we'll isolate the exponentiated expression by adding 2 to each side:
%28x%5E2-x-4%29%5E%283%2F4%29=8
Now we want to get rid of that exponent. Actually evyerthing has an exponent. But we don't see the exponent if it is a 1. So when we say "get rid of an exponent", we really mean "turn the exponent into a 1". So how can we turn the 3/4 into a one? Any number (except 0) can be turned into a 1 by multiplying it by its reciprocal. So if we can figure out an operation that will allow us to multiply the exponent, 3/4, by its reciprocal, 4/3 then the exponent will "disappear". Fortunately we have a rule that says if you raise a power to a power then multiply the exponents. So all we need to do is raise both sides of the equation to the 4/3 power!
%28%28x%5E2-x-4%29%5E%283%2F4%29%29%5E%284%2F3%29=%288%29%5E%284%2F3%29
On the left the exponent turns into a 1 and disappears. On the right we need to simplify 8%5E%284%2F3%29. From what we should know about fractional exponents: 8%5E%284%2F3%29+=+root%283%2C+8%5E4%29+=+%28root%283%2C+8%29%29%5E4. Since 8 is a perfect cube I will use the second expression. (Don't worry, the first expression still works. But it is harder to simplify.) So now our equation is:
x%5E2-x-4+=+%28root%283%2C+8%29%29%5E4
x%5E2-x-4+=+%282%29%5E4
x%5E2-x-4+=+16
At this point we have a quadratic equation to solve. So we want one side to be zero. Subtracting 16 from each side will work:
x%5E2-x-20+=+0
Now we factor (or use the Quadratic Formula). This factors fairly easily:
%28x%2B4%29%28x-5%29+=+0
By the Zero Product property we know that this (or any) product can be zero only if one (or more) of the factors is zero. So:
x%2B4+=+0 or x-5+=+0
Solving these we get:
x+=+-4 or x+=+5

Now we'll check our answers, using the original equation:
%28x%5E2-x-4%29%5E%283%2F4%29-2=6
Checking x = -4:
%28%28-4%29%5E2-%28-4%29-4%29%5E%283%2F4%29-2=6
%2816-%28-4%29-4%29%5E%283%2F4%29-2=6
%2816%2B4-4%29%5E%283%2F4%29-2=6
%2816%29%5E%283%2F4%29-2=6
%28root%284%2C+16%29%29%5E3+-+2+=+6
%282%29%5E3+-+2+=+6
8+-+2+=+6 Check!
Checking x = 5:
%28%285%29%5E2-%285%29-4%29%5E%283%2F4%29-2=6
%2825-%285%29-4%29%5E%283%2F4%29-2=6
%2816%29%5E%283%2F4%29-2=6
%28root%284%2C+16%29%29%5E3+-+2+=+6
%282%29%5E3+-+2+=+6
8+-+2+=+6 Check!

So there are two solutions to the original equation: x = -4 and x = 5.