First, the answer is not 2.
One way to solve for x is to find a way to turn the equation into a polynomial equation (where the exponents are all whole numbers). To start we will rewrite the square root with a fractional exponent:
We can turn the two fractional exponents into whole numbers by raising both sides to an appropriate power. The power that will work is the lowest common denominator (LCD) of the two exponents' denominators, which is 6. Raising both sides to the 6th power we get:
On both sides the rule for the exponents is to multiply them (since they are powers of a power):
As you can see the fractional exponents are gone and we have a polynomial equation.
Next we simplify. We can use the pattern to square the left side quickly:
which simplifies to:
To solve this polynomial we want one side to be zero. After that we will try factoring. Since factoring is easier if the leading coefficient is positive I am going to subtract the entire left side of the equation from both sides giving us:
Now we factor. The greatest common factor is 1 (which we rarely bother factoring out). There are too many terms for any of the factoring patterns or for trinomial factoring. Since I don't see a way to factor by grouping we are left with factoring by trial and error of the possible rational roots.
The possible rational roots of a polynomial are all the ratios, positive and negative, which can be formed using a factor of the constant term (at the end) over a factor of the leading coefficient (at the front). The constant term in this polynomial is 16. (Actually it is -16 but since we use all possible positive and negative ratios it doesn't matter if we use 16 or -16.) The factors of 16 are 1, 2, 4, 8 and 16. The leading coefficient is 1 (whose factors are just 1's). This makes the possible rational roots:
+1/1, +2/1, +4/1, +8/1 and +16/1
which simplify to:
+1, +2, +4, +8 and +16
Checking 1 and -1 are fairly easy to check in your head since powers of 1 and -1 are simple. Neither of these work. The easiest way to see if any of rest are rational roots is to use synthetic division (which I hope you've learned). Trying 2:
2 | 1 -1 -8 -16
--- 2 2 -12
------------------
1 1 -6 -28The remainder is in the lower right corner. Since it is not zero, x-2 is not a factor and 2 is not a root.
Trying 4:
4 | 1 -1 -8 -16
--- 4 12 16
------------------
1 3 4 0The remainder is zero so x-4 is a factor and and 4 is a root. The rest of the bottom row tells us what the other factor is. The "1 3 4" translates into . So factors into:
. The second factor is a quadratic which will not factor further. Not only that the Quadratic Formula will show that its only roots are complex numbers. So assuming you are interested only in real solutions to the equation, there is only one: 4.
Now we check. This is not optional! Earlier we raised both sides of the equation by an even power, 6. Whenever this is done, solutions that are called extraneous solutions may occur. (Extraneous solutions are solutions that fit the equation after it was raised to an even power but do not fit the original equation.) Any extraneous solutions must be rejected and the only way to know if a solution is extraneous is to check. Use the original equation to check:
Checking x = 4:
Simplifying...
Since 1/3 as an exponent means cube root and since , the left side simplifies to a 2 (as does the right side):
Check!!
So the only real solution to your equation is: x = 4.