I have a problem that I am stuck on. I have been asked to solve the problem by finding all real solutions. The problem is: x^6-9x^3+8=0

I thought that I should use the ax^2+bx+c=0 but the 6th power is throwing me off.

If anyone could help, I would really appreciate it!

Thanks,
Ashley

2. Hi

You are right
$x^6-9x^3+8=0$

Let $y = x^3$ then $y^2 = (x^3)^2 = x^6$

And the equation becomes $y^2-9y+8=0$

Solve for y and then find x

3. Hi! Thanks for the help!

The only question I have is- After doing the quadratic formula, is it possible to have to answers for "y"? And if I do have to answers (I have y=3 and y=-4) how do I plug that in for "x"?

Thanks again! Sorry if that is confusing!

4. Originally Posted by aperkin2
I have y=3 and y=-4

You must solve $y^2-9y+8=0$

5. Oh, probably not. After doing the quadratic formula I ended up with--

y= -1 + and - the square root of 49 all over 2
So I thought that would be -8/2 and 6/2
That would be -4 and 3.

I have messed up somewhere I am sure. Thanks again!

6. The solutions are given by the formula

$y = \frac{-b \pm \sqrt{\Delta}}{2a}$

Therefore $y = \frac{9 \pm \sqrt{49}}{2}$

7. Oh I see where I messed up! I was plugging in the wrong value for b. Thank you so much for your help! It has been great!