solve in R:

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- Jul 5th 2013, 12:31 AMdhiabno solution
- Jul 5th 2013, 12:40 AMchiroRe: no solution
Hey dhlab.

What have you tried?

Hint: Try using the rational roots theorem to see if you can get a possibility of a rational root first off. - Jul 5th 2013, 12:41 AMProve ItRe: no solution
Write it as $\displaystyle \displaystyle \begin{align*} X^4 - 4X^3 + 8X^2 - 4X + 3 = 0 \end{align*}$ with $\displaystyle \displaystyle \begin{align*} X = x^2 \end{align*}$, which can be solved using the Quartic Formula. Then solve for $\displaystyle \displaystyle \begin{align*} x \end{align*}$...

- Jul 5th 2013, 02:31 AMibduttRe: no solution
- Jul 5th 2013, 06:55 AMProve ItRe: no solution
I didn't say Quadratic Formula, I said Quartic Formula.

I suggest you read this: - Jul 5th 2013, 10:50 AMdhiabRe: no solution
I suggest this solution

http://www.mathmontada.net/vb/upload...373049942.docx - Jul 5th 2013, 04:22 PMHallsofIvyRe: no solution
What do you

**mean**by saying "$\displaystyle (1- x^2)^4\ne 0$? It certainly**is**equal to 0 or some values of x.

(You are**correct**that $\displaystyle 2x^4+ 2$ is never 0 and that is all you need to show that the sum is not 0.) - Jul 5th 2013, 07:48 PMibduttRe: no solution
by using x = y = B/4 we get

( y+1)^4 - 4 ( y+1)^3 + 8(y+1)^2 + 3 = 0

y^4+2y^2 +8y+8=0

(y^2+1)^2 = 8y -7

How do we proceed from hereonward