MHB How to Solve the Radical Equation Using Fifth Root and Square Root?

[mathdad]

Let frt = fifth root

Let sqrt = square root

Show that frt{176 + 80sqrt{5}} = 1 + sqrt{5}

Do I raise both sides to the 5th power?

Can someone get me started?

 

[mathmari]

Yes. By raising both sides to the 5th power we get:
$$\left (\sqrt[5]{176+80\sqrt{5}}\right )^5=\left (1+\sqrt{5}\right )^5 \Rightarrow 176+80\sqrt{5}=\left (1+\sqrt{5}\right )^5$$

So, we have to calculate $\left (1+\sqrt{5}\right )^5$ and show that it is equal to $176+80\sqrt{5}$:

We have that $$\left (1+\sqrt{5}\right )^5=\left (1+\sqrt{5}\right )^2\cdot \left (1+\sqrt{5}\right )^2\cdot \left (1+\sqrt{5}\right )$$

the first term is:
$$\left (1+\sqrt{5}\right )^2=1+2\cdot \sqrt{5}+5=6+2\sqrt{5}$$

The first two terms are:
$$\left (1+\sqrt{5}\right )^2\cdot \left (1+\sqrt{5}\right )^2=\left (\left (1+\sqrt{5}\right )^2\right )^2=\left (6+2\sqrt{5}\right )^2=36+2\cdot 6\cdot 2\sqrt{5}+4\cdot 5=36+24\sqrt{5}+20 =56+24\sqrt{5}$$

The whole expression is:
\begin{align*}\left (1+\sqrt{5}\right )^2\cdot \left (1+\sqrt{5}\right )^2\cdot \left (1+\sqrt{5}\right )&=\left (\left (1+\sqrt{5}\right )^2\cdot \left (1+\sqrt{5}\right )^2\right )\cdot \left (1+\sqrt{5}\right ) \\ & =\left (56+24\sqrt{5}\right )\cdot \left (1+\sqrt{5}\right ) \\ & =56+24\sqrt{5}+56\sqrt{5}+24\cdot 5 \\ & =56+80\sqrt{5}+120 \\ & =176+80\sqrt{5}\end{align*}

 

[MarkFL]

An alternate approach would be to use the binomial theorem:

$$(1+\sqrt{5})^5=\sum_{k=0}^{5}\left({5 \choose k}\sqrt{5}^k\right)$$

$$(1+\sqrt{5})^5=1+5\sqrt{5}+50+50\sqrt{5}+125+25\sqrt{5}=176+80\sqrt{5}$$

And so:

$$\sqrt[5]{176+80\sqrt{5}}=\sqrt[5]{(1+\sqrt{5})^5}=1+\sqrt{5}$$

 

[mathdad]

Yes. By raising both sides to the 5th power we get:
$$\left (\sqrt[5]{176+80\sqrt{5}}\right )^5=\left (1+\sqrt{5}\right )^5 \Rightarrow 176+80\sqrt{5}=\left (1+\sqrt{5}\right )^5$$

So, we have to calculate $\left (1+\sqrt{5}\right )^5$ and show that it is equal to $176+80\sqrt{5}$:

We have that $$\left (1+\sqrt{5}\right )^5=\left (1+\sqrt{5}\right )^2\cdot \left (1+\sqrt{5}\right )^2\cdot \left (1+\sqrt{5}\right )$$

the first term is:
$$\left (1+\sqrt{5}\right )^2=1+2\cdot \sqrt{5}+5=6+2\sqrt{5}$$

The first two terms are:
$$\left (1+\sqrt{5}\right )^2\cdot \left (1+\sqrt{5}\right )^2=\left (\left (1+\sqrt{5}\right )^2\right )^2=\left (6+2\sqrt{5}\right )^2=36+2\cdot 6\cdot 2\sqrt{5}+4\cdot 5=36+24\sqrt{5}+20 =56+24\sqrt{5}$$

The whole expression is:
\begin{align*}\left (1+\sqrt{5}\right )^2\cdot \left (1+\sqrt{5}\right )^2\cdot \left (1+\sqrt{5}\right )&=\left (\left (1+\sqrt{5}\right )^2\cdot \left (1+\sqrt{5}\right )^2\right )\cdot \left (1+\sqrt{5}\right ) \\ & =\left (56+24\sqrt{5}\right )\cdot \left (1+\sqrt{5}\right ) \\ & =56+24\sqrt{5}+56\sqrt{5}+24\cdot 5 \\ & =56+80\sqrt{5}+120 \\ & =176+80\sqrt{5}\end{align*}

Thank you very much for providing such a detailed reply.

- - - Updated - - -

An alternate approach would be to use the binomial theorem:

$$(1+\sqrt{5})^5=\sum_{k=0}^{5}\left({5 \choose k}\sqrt{5}^k\right)$$

$$(1+\sqrt{5})^5=1+5\sqrt{5}+50+50\sqrt{5}+125+25\sqrt{5}=176+80\sqrt{5}$$

And so:

$$\sqrt[5]{176+80\sqrt{5}}=\sqrt[5]{(1+\sqrt{5})^5}=1+\sqrt{5}$$

Thanks a million. I had no idea that the binomial theorem is somehow connected here. I learn something new every time I visit this site.