# Finding the roots and K of a quartic.

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• December 25th 2007, 11:02 PM
amardeep33
Finding the roots and K of a quartic.[solved]
Find all roots and the value of k in the equation:
2x^4 + 3x^3 - kx^2 - 48x + 32 = 0.?
Given that the sum of 2 of the roots = 0.
I've been trying for half an hour, and I keep winding up with an equation with two variables.
I know there are 4 roots, R1, R2, R3, R4.
I said R1= R1 || R2= -R1 || R3= R3 and R4=R4.

I know S1(Sum of roots taken one at a time) = -3/2 = R3 + R4
S3 = 24 = (-R1^2 * R3) - (R1^2 * R4) - (R1R3R4)
S4 = 16 = -R1^2 * R3 * R4
What I did first was multiply the S4 eqn by 3/2, to make it = to 24. Then I set the new S4 and the S3 eqns equal to each other and worked from there. I keep getting something unsolvable.
Any help is greatly appreciated, but it'd be awesome if you could give me some steps or pointers on where I might've gone wrong.
Thanks alot in advance.

Solved. Pretty easy actually.
• December 25th 2007, 11:51 PM
kalagota
Quote:

Originally Posted by amardeep33
Find all roots and the value of k in the equation:
2x^4 + 3x^3 - kx^2 - 48x + 32 = 0.?
Given that the sum of 2 of the roots = 0.
I've been trying for half an hour, and I keep winding up with an equation with two variables.
I know there are 4 roots, R1, R2, R3, R4.
I said R1= R1 || R2= -R1 || R3= R3 and R4=R4.

I know S1(Sum of roots taken one at a time) = -3/2 = R3 + R4
S3 = 24 = (-R1^2 * R3) - (R1^2 * R4) - (R1R3R4)
S4 = 16 = -R1^2 * R3 * R4
What I did first was multiply the S4 eqn by 3/2, to make it = to 24. Then I set the new S4 and the S3 eqns equal to each other and worked from there. I keep getting something unsolvable.
Any help is greatly appreciated, but it'd be awesome if you could give me some steps or pointers on where I might've gone wrong.
Thanks alot in advance.

Solved. Pretty easy actually.

$2x^4 + 3x^3 - kx^2 - 48x + 32 = x^4 + \frac{3}{2}x^3 - \frac{k}{2}x^2 - 24x + 16 = 0$

$x^4 + \frac{3}{2}x^3 - \frac{k}{2}x^2 - 24x + 16 = (x-a)(x+a)(x-b)(x-c) =0$, where $a,-a,b,c$ are the roots..

$(x-a)(x+a)(x-b)(x-c) = x^4 + (-b-c)x^3 - (a^2 - bc)x^2 - (-a^2b - a^2c)x - a^2bc = 0$

thus, solve for the system

$\left\{ {\begin{array}{ccl} -b-c & = & \frac{3}{2} \\ \, & \, & \, \\ a^2 - bc & = & \frac{k}{2} \\ \, & \, & \, \\ -a^2b - a^2c & = & 24 \\ \, & \, & \, \\ a^2bc & = & -16 \\ \end{array}} \right.$

that should be easier.. Ü

i didn't notice, you've solved it already..