Rationalize the denominator of $\frac{\sqrt{32}}{\sqrt{16}-\sqrt{2 }}$. The answer can be written as $\frac{A\sqrt{B}+C}{D}$, where $A$, $B$, $C$, and $D$ are integers, $D$ is positive, and $B$ is not divisible by the square of any prime. Find the minimum possible value of $A+B+C+D$.

Rationalizing the denominator involves exploiting the well-known difference of squares formula,

We have

so that

Rewrite 16 and 32 as powers of 2, then simplify:

So we have A = 8, B = 2, C = 4, and D = 7, and thus A + B + C + D = 21.

option: d is correct

step-by-step explanation:

step-by-step explanation:

this can be solved first with pythagorean theorem, which tells us to square both legs of the triangle and sum it. this will be equal to the square of the hypotenuse.

the hypotenuse is the side "opposite" the 90 degree angle, which is x.

so we can write using pythagorean theorem,

simplifying:

to further simplify this radical, we need to use the property  

thus