# Question 4824d

Aug 18, 2017

${x}^{2} + \frac{16}{x} ^ 2 = 28$

#### Explanation:

The first thing to notice here is that

${x}^{2} + \frac{16}{x} ^ 2$

can be written as

${\left(x + \frac{4}{x}\right)}^{2} - 8$

This is the case because

${\left(x + \frac{4}{x}\right)}^{2} = {x}^{2} + 2 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} \cdot \frac{4}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}} + {\left(\frac{4}{x}\right)}^{2}$

$= {x}^{2} + 8 + \frac{16}{x} ^ 2$

So if you subtract $8$ from both sides of this equation, you will end up with

${\left(x + \frac{4}{x}\right)}^{2} - 8 = {x}^{2} + \textcolor{red}{\cancel{\textcolor{b l a c k}{8}}} + \frac{16}{x} ^ 2 - \textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}$

which is

(x + 4/x)^2-8 = x^2 + 16/x^2" "color(blue)("(*)")

Now, you know that

$x = 3 - \sqrt{5}$

This means that you have

$x + \frac{4}{x} = 3 - \sqrt{5} + \frac{4}{3 - \sqrt{5}}$

If you rationalize the denominator of $\frac{4}{3 - \sqrt{5}}$ by multiplying both the numerator and the denominator by $3 + \sqrt{5}$, the conjugate of $3 - \sqrt{5}$, you will end up with

3 - sqrt(5) + (4 * (3 + sqrt(5)))/((3 - sqrt(5))(3 + sqrt(5))#

At this point, you can use the fact that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)}}}$

to say that

$\left(3 - \sqrt{5}\right) \left(3 + \sqrt{5}\right) = {3}^{2} - {\left(\sqrt{5}\right)}^{2}$

$= 9 - 5$

$= 4$

This means that you have

$3 - \sqrt{5} + \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} \cdot \left(3 + \sqrt{5}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}} = 3 - \textcolor{red}{\cancel{\textcolor{b l a c k}{\sqrt{5}}}} + 3 + \textcolor{red}{\cancel{\textcolor{b l a c k}{\sqrt{5}}}}$

$= 6$

Therefore, you can say that

$x + \frac{4}{x} = 6$

Plug this back into equation $\textcolor{b l u e}{\text{(*)}}$ to get

${\left(x + \frac{4}{x}\right)}^{2} - 8 = {6}^{2} - 8 = 28$

This means that you have

${x}^{2} + \frac{16}{x} ^ 2 = 28$