Question 345185
{{{(x^2-x-4)^(3/4)-2=6}}}
First we'll isolate the exponentiated expression by adding 2 to each side:
{{{(x^2-x-4)^(3/4)=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!
{{{((x^2-x-4)^(3/4))^(4/3)=(8)^(4/3)}}}
On the left the exponent turns into a 1 and disappears. On the right we need to simplify {{{8^(4/3)}}}. From what we should know about fractional exponents: {{{8^(4/3) = root(3, 8^4) = (root(3, 8))^4}}}. 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^2-x-4 = (root(3, 8))^4}}}
{{{x^2-x-4 = (2)^4}}}
{{{x^2-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^2-x-20 = 0}}}
Now we factor (or use the Quadratic Formula). This factors fairly easily:
{{{(x+4)(x-5) = 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+4 = 0}}} or {{{x-5 = 0}}}
Solving these we get:
{{{x = -4}}} or {{{x = 5}}}<br>
Now we'll check our answers, using the original equation:
{{{(x^2-x-4)^(3/4)-2=6}}}
Checking x = -4:
{{{((-4)^2-(-4)-4)^(3/4)-2=6}}}
{{{(16-(-4)-4)^(3/4)-2=6}}}
{{{(16+4-4)^(3/4)-2=6}}}
{{{(16)^(3/4)-2=6}}}
{{{(root(4, 16))^3 - 2 = 6}}}
{{{(2)^3 - 2 = 6}}}
{{{8 - 2 = 6}}} Check!
Checking x = 5:
{{{((5)^2-(5)-4)^(3/4)-2=6}}}
{{{(25-(5)-4)^(3/4)-2=6}}}
{{{(16)^(3/4)-2=6}}}
{{{(root(4, 16))^3 - 2 = 6}}}
{{{(2)^3 - 2 = 6}}}
{{{8 - 2 = 6}}} Check!<br>
So there are two solutions to the original equation: x = -4 and x = 5.