## mathtard Group Title w^4-14w^2-2=0 solve 2 years ago 2 years ago

1. satellite73 Group Title

gimmick is to put $x=w^2$ and write $x^2-14x-2=0$ solve for x

2. satellite73 Group Title

then go back. solve for x via $x^2-14x=2$ $(x-7)^2=2+49$ $(x-7)^2=51$ $x-7=\pm\sqrt{51}$ $x=7\pm\sqrt{51}$

3. satellite73 Group Title

now $7-\sqrt{51}<0$ so you cannot have $w^2=7-\sqrt{51}$ with real numbers. therefore $w^2=7+\sqrt{51}$ and finally $w=\pm\sqrt{7+\sqrt{51}}$

4. mathtard Group Title

thank you!

5. satellite73 Group Title

yw