Question

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 $A+B+C+D$.

Answer 1

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

$a^2-b^2=(a-b)(a+b)$

We have

$(\sqrt{16}-\sqrt2)(\sqrt{16}+\sqrt2)=(\sqrt{16})^2-(\sqrt2)^2=16-2=14$

so that

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{\sqrt{32}(\sqrt{16}+\sqrt2)}{14}$

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

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{\sqrt{2^5}(\sqrt{2^4}+\sqrt2)}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{2^2\sqrt2(2^2+\sqrt2)}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{4\sqrt2(4+\sqrt2)}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{16\sqrt2+4(\sqrt2)^2}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{16\sqrt2+8}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{8\sqrt2+4}7$

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