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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 $A+B+C+D$.

Respuesta :

LammettHash

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

[tex]a^2-b^2=(a-b)(a+b)[/tex]

We have

[tex](\sqrt{16}-\sqrt2)(\sqrt{16}+\sqrt2)=(\sqrt{16})^2-(\sqrt2)^2=16-2=14[/tex]

so that

[tex]\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{\sqrt{32}(\sqrt{16}+\sqrt2)}{14}[/tex]

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

[tex]\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{\sqrt{2^5}(\sqrt{2^4}+\sqrt2)}{14}[/tex]

[tex]\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{2^2\sqrt2(2^2+\sqrt2)}{14}[/tex]

[tex]\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{4\sqrt2(4+\sqrt2)}{14}[/tex]

[tex]\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{16\sqrt2+4(\sqrt2)^2}{14}[/tex]

[tex]\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{16\sqrt2+8}{14}[/tex]

[tex]\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{8\sqrt2+4}7[/tex]

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

The rationalized form of the given surd is [tex]\frac{8\sqrt{2}+4}{7}[/tex], and the minimum possible value of A + B + C + D = 8 + 2 + 4 + 7 = 21

Rationalizing Surds

From the question, we are to rationalize the denominator of the given surd.

We are to write the answer in the form

[tex]\frac{A\sqrt{B}+C}{D}[/tex]

and find the minimum possible value of A + B + C + D

The given surd is

[tex]\frac{\sqrt{32}}{\sqrt{16}-\sqrt{2}}[/tex]

To rationalize the surd, we will multiply the numerator and denominator by the conjugate of the denominator

The conjugate of the denominator is [tex]\sqrt{16}+\sqrt{2}[/tex]

Therefore,

[tex]\frac{\sqrt{32}}{\sqrt{16}-\sqrt{2}} \times \frac{\sqrt{16}+\sqrt{2}}{\sqrt{16}+\sqrt{2}}[/tex]

[tex]= \frac{\sqrt{32}(\sqrt{16}+\sqrt{2})}{(\sqrt{16}-\sqrt{2})(\sqrt{16}+\sqrt{2})}[/tex]

[tex]= \frac{\sqrt{512}+\sqrt{64})}{(\sqrt{16})^{2} -(\sqrt{2})^{2} }[/tex]

[tex]= \frac{16\sqrt{2}+8}{16-2}[/tex]

[tex]= \frac{16\sqrt{2}+8}{14}[/tex]

[tex]= \frac{2(8\sqrt{2}+4)}{2(7)}[/tex]

By comparing with, [tex]\frac{A\sqrt{B}+C}{D}[/tex]

A = 8, B = 2, C = 4, and D = 7

Then, the minimum possible value of A + B + C + D = 8 + 2 + 4 + 7 = 21

Hence, the rationalized form of the given surd is [tex]\frac{8\sqrt{2}+4}{7}[/tex], and the minimum possible value of A + B + C + D = 8 + 2 + 4 + 7 = 21

Learn more on Rationalizing surds here: https://brainly.com/question/9547165

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