Solving for k and m in a Quartic Equation

[olee]

Homework Statement

The equation x⁴-6x³-73x²+kx+m=0 has two positive roots, \alpha \beta, and two negative roots, \delta \gamma. It is given that \alpha\beta=\delta\gamma=4

(i) Find the values of k and m

Homework Equations

stated above

The Attempt at a Solution

m = (\alpha\beta)(\gamma\delta)=4x4=16

not too sure how to approach solving k

 

[epenguin]

You can at least take the natural step and then consider.

You got your result by comparing m with the constant term in the expansion of (x - \alpha)(x - \beta)(x-\gamma)(x - \delta) .

So you should be able to write what coefficient of x is in the expansion (+ has probably been covered in your course) so at least write that down and see if there is anything you can do with it.

(As far as I can see this comes out a whole number, but the equation itself does not have any nice whole number solutions.)

 

[snshusat161]

It seems that you are confused, I'm giving the relations between the roots and coefficients in such equation

First make the coefficient of highest term as unity, in your case it is already one.

Now,

(1). Second term coefficient is equal to the sum of roots but with its sign changed.

Note: For simplicity I'm writing roots as A, B, C and D

so here, A + B + C + D = -(-6)
= 6 ......(equation 1)

(2) Third coefficient is equal to the product of roots taken two at the time

so here AB + BC + CD + DA = -73 ...... (No use for you in this problem)

(3) Fourth Coefficient (which you have to find "k") is equal to the product of roots taken three at the time with its sign changed.

so here

ABC + BCD + CDA + DAB = - k
Given AB = CD = 4

4C + 4B + 4A + 4D = -k

4 (A + B + C + D) = -k

From equation (1)

4 (6) = -k

k = -24 .... (your first answer)

(4) and last coefficient (constant therm here ) is equal to the sum of roots taken four at a time

so here

ABCD = m

given AB = CD =4

so 4 (4) = m

m = 16...... (your second answer)