# Need help solving equations.

• Sep 24th 2010, 06:14 AM
nrub87
Need help solving equations.
$(x^4-1)^6-5(x^4-1)^3+6=0$ leave answer in radical form. Show all work please.

solve : x-5(sqrtx)+6=0 by factoring left side directly (no substitution)

solve : same question but this time isolate the radical and then squaring each side

Thank you so much
• Sep 24th 2010, 06:20 AM
emakarov
For the first question, you can denote t = (x^4 - 1)^3. Then you will get a quadratic equation in t.

For the rest, what have you tried and what difficulties are you having?
• Sep 24th 2010, 06:56 AM
nrub87
i got the first one now.. thanks!
But im totaly stuck on the other to. I don't know how to begin, or continue for that matter! ;)
• Sep 24th 2010, 07:42 AM
emakarov
Quote:

x-5(sqrtx)+6
You can factor this, for example, by completing the square. Namely, $x-5\sqrt{x}+6=(\sqrt{x}-5/2)^2-\dots$. Find what must go instead of ... and represent that number as a square, i.e., $y^2$ for some number $y$. Then $x-5\sqrt{x}+6 = (\sqrt{x}-5/2 - y)(\sqrt{x}-5/2 + y)$.

Quote:

same question but this time isolate the radical and then squaring each side
This tells you what to do. If you have difficulty proceeding, post the result after isolating the radical and then squaring each side.