If $\alpha$ and $\beta$ are the roots of the equation $x^2-34x+1=0$,find the value of $\sqrt[4]{\alpha}-\sqrt[4]{\beta}$,where $\sqrt[4]{.}$ denotes the principal value.
I found out the $\alpha$ and $\beta$.
$\alpha,\beta=\frac{34\pm\sqrt{32\times 36}}{2}=17\pm12\sqrt2$ but i do not know how to find $\sqrt[4]{\alpha}-\sqrt[4]{\beta}$.
On
From the properties of the roots of a quadratic equation we can deduce that:$$\alpha+\beta=34\tag{1}$$$$\alpha\beta=1\tag{2}$$Next we see that:$$(\sqrt[4]{\alpha}-\sqrt[4]{\beta})^2=\sqrt{\alpha}+\sqrt{\beta}-2\sqrt[4]{\alpha\beta}=\sqrt{\alpha}+\sqrt{\beta}-2\text{ (using (2) above)}\tag{3}$$We also see that:$$(\sqrt{\alpha}+\sqrt{\beta})^2=\alpha+\beta+2\sqrt{\alpha\beta}=34+2\text{ (using (1) and (2) above)}$$$$\therefore\sqrt{\alpha}+\sqrt{\beta}=\sqrt{36}=6$$Now substitute this into (3) and hopefully you can finish this off.
One may observe that $$ (\sqrt{2}-1)^2=3-2\sqrt{2} $$ and that $$ (3-2\sqrt{2})^2=17-12\sqrt{2} $$ thus $$ (\sqrt{2}-1)^4=17-12\sqrt{2} $$ giving
similarly