2

I've been wrapping my head around this question lately:

Let

$$a = \frac{9+\sqrt{45}}{2}$$

Find the value of

$$\frac{1}{a}$$

I've done it like this:

$$\frac{1}{a}=\frac{2}{9+\sqrt{45}}$$

I rationalize the denominator like this:

$$\frac{1}{a}=\frac{2}{9+\sqrt{45}} \times (\frac{9-\sqrt{45}}{9-\sqrt{45}})$$

This is what I should get:

$$\frac{1}{a} = \frac{2(9-\sqrt{45})}{81-45} \rightarrow \frac{1}{a}=\frac{18-2\sqrt{45}}{36})$$

Which I can simplify to:

$$\frac{1}{a}=\frac{2(9-\sqrt{45})}{36}\rightarrow\frac{1}{a}=\frac{9-\sqrt{45}}{18}$$

However, this answer can't be found in my multiple choice question here:

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Any hints on what I'm doing wrong?

Cesare
  • 1,471

4 Answers4

4

$$\frac { 1 }{ a } =\frac { 9-\sqrt { 45 } }{ 18 } =\frac { 3\left( 3-\sqrt { 5 } \right) }{ 18 } =\frac { 3-\sqrt { 5 } }{ 6 } $$

haqnatural
  • 21,578
2

The answer is b, which is equivalent to your answer after simplifying.

This is because $ 9 - \sqrt{45} = 9 - \sqrt{9 * 5} = 9 - 3\sqrt{5}$. Then,$ \frac{9 - 3\sqrt{5}}{18} = \frac{3 - \sqrt{5}}{6}. $

Vijay Ramanujan
  • 115
1

Since it is a multiple choice question, you can just multiply the given term with all the possible terms and verify when you get $1$

Peter
  • 84,454
1

Since the $ 45 $ inside the radical is expressed a product containing a perfect square ($ 9 $), $$ \sqrt{45} = \sqrt{9 \times 5} = \sqrt{9}\sqrt{5} = 3\sqrt{5} $$

Thus, $$ \frac{1}{a} = \frac{9 - \sqrt{45}}{18} = \frac{9 - 3\sqrt{5}}{18} $$

Dividing the numerator and denominator by $ 3 $ (the greatest common factor of the coefficients $ 3 $ and $ 9 $ in the numerator and $ 18 $ in the denominator):

$$ \frac{(9 - 3\sqrt{5}) / 3}{18 / 3} = \frac{3 - \sqrt{5}}{6} $$ which is one of the choices (choice (b)).

Marvin
  • 548