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konstantin123 [22]
3 years ago
11

Rationalize the denominator of $\frac{\sqrt{32}}{\sqrt{16}-\sqrt{2}}$. The answer can be written as $\frac{A\sqrt{B}+C}{D}$, whe

re $A$, $B$, $C$, and $D$ are integers, $D$ is positive, and $B$ is not divisible by the square of any prime. Find the minimum possible value of $A+B+C+D$.
Mathematics
1 answer:
musickatia [10]3 years ago
3 0

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 <em>A</em> = 8, <em>B</em> = 2, <em>C</em> = 4, and <em>D</em> = 7, and thus <em>A</em> + <em>B</em> + <em>C</em> + <em>D</em> = 21.

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