Show that $\frac {37 - \sqrt{1357}}{3} = \frac {4}{37 + \sqrt{1357}}$.

Solving the cubic equation:

$$3x^3 - 74x^2 + 4x = 0$$

I found the following roots:

$$x = 0, \frac {37 - \sqrt{1357}}{3}, \frac {37 + \sqrt{1357}}{3}$$

On the Wolfram Alpha website one of the roots is shown differently.

$$x = \frac {37 - \sqrt{1357}}{3}$$

The site shows:

$$x = \frac {4}{37 + \sqrt{1357}}$$

I rapidly established that:

$$\frac {37 - \sqrt{1357}}{3} = \frac {4}{37 + \sqrt{1357}} \approx 0.054$$

But I can't see what mathematical steps would turn $\frac {37 - \sqrt{1357}}{3}$ into $\frac {4}{37 + \sqrt{1357}}$ or indeed visa-versa. Can someone explain what steps would make this transformation please?

Also...

It seems to me that if an equation has 2 roots which differ only in whether they have a plus or minus sign as a result of taking the square root of both sides of a quadratic equation, then it's best to express the 2 roots in the same way with only the plus-or-minus sign differing (it's clearer that way right?). What are the reasons for expressing one of them differently as Wolfram Alpha has done in this case?

Thanks.

• Are you aware of the conjugates? Sep 24, 2017 at 15:54
• Multiply by conjugate.
– user312097
Sep 24, 2017 at 15:54
• There are more advanced ways of solving equations that rely on finding a one radical quantity and expressing all roots in terms of it (Viète's method for cubics, for example). This may be what Wolfram Alpha is up to, although someone may have to pay to find out. Sep 24, 2017 at 16:12
• Note, for example, that the two solutions to $x^2+2bx+c=0$ can be written as $\frac{1}{2}(b+\sqrt{b^2-c})$ and $\frac{c}{2}\frac{1}{b+\sqrt{b^2-c}}$. This example is also nice because you get extra information, in this case the rather trivial fact that if $c=0$, $0$ is a root. Sep 24, 2017 at 22:55
• That should be $-b+\sqrt{b^2-c}$, of course. Sep 24, 2017 at 23:09

$$\frac {37 - \sqrt{1357}}{3}= \frac{(37-\sqrt{1357})(37+\sqrt{1357})}{3(37+\sqrt{1357})}=\frac{37^2-1357}{3(37+\sqrt{1357})}$$ $$=\frac{12}{3(37+\sqrt{1357})}=\frac{4}{(37+\sqrt{1357})}$$
Just note that $(37-\sqrt{1357})(37+\sqrt{1357})=37^2-1357=12=3\times4$. Therefore$$\frac{37-\sqrt{1357}}3=\frac4{37+\sqrt{1357}}.$$