Rationalising the denominator in $\frac8{\sqrt5+1}$

Question

Given this expression

$\displaystyle{8 \over {\sqrt 5 + 1}}$

I multiply the nominator and denominator by the conjugate:

$\displaystyle{{8 \over {\sqrt 5 + 1}} \times {{\sqrt 5 - 1}\over{\sqrt 5 -1}}}$

$\displaystyle={{8\sqrt 5 - 8} \over {\sqrt 25 - \sqrt 5 + \sqrt 5 - 1}}$

$\displaystyle={{8 \sqrt 5 - 8} \over 4}$

$={{2 \sqrt 5} -8}$

But the answer in the textbook is:

${2 \sqrt 5 -2}$

I can't see the discrepancy on my end.

Answer 1

You have done it correct upto the last step where you made a mistake, $$\frac{8\sqrt5-8}4=4\cdot\frac{2\sqrt5-2}4=2\sqrt5-2$$

This is because when you divide something like this:

$$\frac{4\times2+4\times4}4,$$

You cant cancel the 4 from a single term, you have to cancel it from both,

$$\frac{4\times2+4\times4}4\neq2+16$$ $$\frac{4\times2+4\times4}4=\frac{4(2+4)}4=2+4.$$

Answer 2

We can write $\displaystyle \frac{8}{\sqrt{5}+1} = \frac{8}{\sqrt{5}+1} \times \frac{\left(\sqrt{5}-1\right)}{\left(\sqrt{5}-1\right)} = \frac{8(\sqrt{5}-1)}{(\sqrt{5})^2-(1)^2} = \frac{8(\sqrt{5}-1)}{4}= 2\sqrt{5}-2$

Answer 3

$$\frac { 8 }{ \sqrt { 5 } +1 } \cdot \frac { \sqrt { 5 } -1 }{ \sqrt { 5 } -1 } =\frac { 8\left( \sqrt { 5 } -1 \right) }{ { \left( \sqrt { 5 } \right) }^{ 2 }-1^{ 2 } } =\frac { 8\left( \sqrt { 5 } -1 \right) }{ 4 } =2\left( \sqrt { 5 } -1 \right) $$

Answer 4

\begin{align} \ {{8 \over {\sqrt 5 + 1}} \times {{\sqrt 5 - 1}\over{\sqrt 5 -1}}} \\ \ 8 \times {(\sqrt 5 - 1)} \over {(\sqrt 5 + 1)\times {(\sqrt 5 - 1)}} \\ \ 8 \times \sqrt 5 - 8 \over {(\sqrt 5)^2 - 1^2} \\ \ 8 \times \sqrt 5 - 8 \over {4} \\ \ 2 \times \sqrt 5 - 2 \end{align}

Answer 5

As stated by @Kartik, everything is correct but the last step. Here you have:

$$ \frac{8 \sqrt 5 - 8}{4}= \frac{8 \sqrt 5}{4} - \color{red}{\frac{8}{4}}= 2 \sqrt 5 - \color{red}{2}. $$