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Homework Statement
If the sum of two roots of:
[tex]x^4+2x^3-8x^2-18x-9=0[/tex]
are equal to 0, find the roots of the equation.
Homework Equations
For this quartic, [tex]P(x)=ax^4+bx^3+cx^2+dx+e[/tex]
Let the roots be [tex]\alpha, \beta, \gamma, \delta[/tex]
[tex]\alpha +\beta +\gamma +\delta=\frac{-b}{a}[/tex]
[tex]\alpha\beta +\alpha\gamma +\alpha\delta +\beta\gamma +\beta\delta +\gamma\delta=\frac{c}{a}[/tex]
[tex]\alpha\beta\gamma +\alpha\beta\delta +\beta\gamma\delta=\frac{-d}{a}[/tex]
[tex]\alpha\beta\gamma\delta=\frac{e}{a}[/tex]
The Attempt at a Solution
Since 2 roots are equal to 0, i.e. [tex]\alpha +\beta=0[/tex] therefore, [tex]\beta =-\alpha[/tex]
So I began to rearrange the equations, trying to find values for each root by making it the subject. When I end up 'simplifying' the equations down to 2 simultaneous equations, with 2 roots, trying to make the subject for one becomes chaotic.
Let [tex]\beta =-\alpha[/tex]
Therefore, by the sum of the roots one at a time:
[tex]\alpha -\alpha +\beta +\delta=-2[/tex]
Thus, [tex]\beta = -(2+\delta )[/tex]
Now the roots are: [tex]\alpha ,-\alpha ,\delta ,-(\delta +2)[/tex]
By multiplying roots two at a time:
[tex]-\alpha ^2 + \alpha \delta -\alpha (\delta +2) -\alpha\delta +\alpha(\delta +2)-\delta (\delta +2)=-8[/tex]
Simplified: [tex]\alpha ^2+\delta ^2-2\delta =8[/tex] (1)
By multiplying three at a time:
[tex]-\alpha ^2\delta +\alpha ^2(\delta +2) +\alpha \delta (\delta +2)=18[/tex]
Simplified: [tex]\alpha (2\alpha +\delta ^2+2\delta )=18[/tex] (2)
By multiplying all at a time:
[tex]\alpha ^2\delta (\delta +2)=-9[/tex] (3)
Now with equations (1), (2) and (3). It is possible to find [tex]\alpha[/tex] or [tex]\delta[/tex] by making them the subject through simultaneous equations, but this becomes ridiculously complicated which just looks like another polynomial.
Is there another method to solve this problem, or is the answer to the roots easier than meets the eye?
p.s. I know I can just graph the polynomial to find the roots that way, since they're integral, but I *think* it's meant to be solved somewhat along these lines.