# no solution

• Jul 5th 2013, 12:31 AM
dhiab
no solution
• Jul 5th 2013, 12:40 AM
chiro
Re: 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 AM
Prove It
Re: 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 AM
ibdutt
Re: no solution
Could you please guide further as to how do we use quadratic formula for a bi-quadratic equation.
thanks

Quote:

Originally Posted by Prove It
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, 06:55 AM
Prove It
Re: no solution
I didn't say Quadratic Formula, I said Quartic Formula.

• Jul 5th 2013, 10:50 AM
dhiab
Re: no solution
• Jul 5th 2013, 04:22 PM
HallsofIvy
Re: 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 PM
ibdutt
Re: 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