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# help how to rationalize denominator

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Rationalize the denominator of $\displaystyle \frac{1}{2 - \sqrt{2}}$. With your answer in the form $\displaystyle \frac{\sqrt{A} + \sqrt{B} + \sqrt{C}}{D}$, and the fraction in lowest terms, what is $A + B + C + D$?

Aug 18, 2023

#1
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To rationalize an expression such as $$\frac{1}{2 - \sqrt{2}}$$, you first recognize that there is a cube root in the denominator. This immediately provides insight as to how to approach this particular problem. Personally, I find working with fractional exponents is far easier than radicals, so I will almost always convert to this.

Since there is a cube root within the denominator, we will have to make the denominator a perfect cube in order to ensure that the denominator has no radicals. We can do this by making a key observation: $$A^3 - B^3 = (A - B)(A^2 + AB + B^2)$$. This is the formula for the difference of a perfect cube. This information should be enough to rationalize this problem and ones like it.

If we look closely at the denominator, we can match up the formula for a perfect cube to the denominator of this particular problem.

If we let $$A = 2$$ and $$B = \sqrt{2} = 2^\frac{1}{3}$$, then that would make the following identity.

$$A = 2; B = 2^\frac{1}{3} \\ 2^3 - \left(2^\frac{1}{3}\right)^3 = \left(2 - 2^\frac{1}{3}\right)\left(2^2 + 2*2^\frac{1}{3} + \left(2^\frac{1}{3}\right)^2\right) \\ 6 = \left(2 - 2^\frac{1}{3} \right) \left(4 + 2^\frac{4}{3} + 2^\frac{2}{3}\right) \\$$

Now, we can rationalize because we have the term that will cancel out all radicals within the denominator.

$$\frac{1}{2 - 2^\frac{1}{3}} * \frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{4 + 2^\frac{4}{3} + 2 ^\frac{2}{3}} = \frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{6}$$

Now, we have to convert to the radical notation and figure out how it matches the desired format of $$\displaystyle \frac{\sqrt{A} + \sqrt{B} + \sqrt{C}}{D}$$.

$$\frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{6} = \frac{4 + \sqrt{2^4} + \sqrt{2^2}}{6} = \frac{\sqrt{64} + \sqrt{16} + \sqrt{4}}{6}$$

Now, we identify the desired variables and find their sum.

$$A = 64, B=16, C=4, D=6 \\ A + B + C + D = 64 + 16 + 4 + 6 = 90$$

Aug 19, 2023

#1
+1

To rationalize an expression such as $$\frac{1}{2 - \sqrt{2}}$$, you first recognize that there is a cube root in the denominator. This immediately provides insight as to how to approach this particular problem. Personally, I find working with fractional exponents is far easier than radicals, so I will almost always convert to this.

Since there is a cube root within the denominator, we will have to make the denominator a perfect cube in order to ensure that the denominator has no radicals. We can do this by making a key observation: $$A^3 - B^3 = (A - B)(A^2 + AB + B^2)$$. This is the formula for the difference of a perfect cube. This information should be enough to rationalize this problem and ones like it.

If we look closely at the denominator, we can match up the formula for a perfect cube to the denominator of this particular problem.

If we let $$A = 2$$ and $$B = \sqrt{2} = 2^\frac{1}{3}$$, then that would make the following identity.

$$A = 2; B = 2^\frac{1}{3} \\ 2^3 - \left(2^\frac{1}{3}\right)^3 = \left(2 - 2^\frac{1}{3}\right)\left(2^2 + 2*2^\frac{1}{3} + \left(2^\frac{1}{3}\right)^2\right) \\ 6 = \left(2 - 2^\frac{1}{3} \right) \left(4 + 2^\frac{4}{3} + 2^\frac{2}{3}\right) \\$$

Now, we can rationalize because we have the term that will cancel out all radicals within the denominator.

$$\frac{1}{2 - 2^\frac{1}{3}} * \frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{4 + 2^\frac{4}{3} + 2 ^\frac{2}{3}} = \frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{6}$$

Now, we have to convert to the radical notation and figure out how it matches the desired format of $$\displaystyle \frac{\sqrt{A} + \sqrt{B} + \sqrt{C}}{D}$$.

$$\frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{6} = \frac{4 + \sqrt{2^4} + \sqrt{2^2}}{6} = \frac{\sqrt{64} + \sqrt{16} + \sqrt{4}}{6}$$

Now, we identify the desired variables and find their sum.

$$A = 64, B=16, C=4, D=6 \\ A + B + C + D = 64 + 16 + 4 + 6 = 90$$

The3Mathketeers Aug 19, 2023
#2
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Very nice, The 3Mathketeers  !!!!!

This one stumped me........your approach and  explanation is excellent  !!!!   CPhill  Aug 19, 2023