Given the equation:$$3x^4-25x^3+50x^2-kx+12=0$$ has two roots whose product is $2$. It was given to find the value of k and all the roots of the equation.This is how I solved it:

Let the roots be $$\alpha , \beta , \gamma , \delta$$ Let $$\alpha \beta=2$$So,$$\gamma \delta=2$$ $$\implies(x^2-(\alpha +\beta)x+2)(x^2-(\gamma +\delta)x+2)=0$$ Then by expanding we get :$$x^4-(\alpha+\beta+\gamma+\delta)x^3+(\alpha \gamma +\beta \gamma + \alpha \delta + \beta \delta +4)x^2-2(\alpha+\beta+\gamma+\delta)x+4=0$$ Comparing coefficients, we get $k=50$

Now,$$\alpha\gamma+\beta\gamma+\alpha\delta+\beta\delta+4=50/3$$ $$\implies (\alpha+\beta)(\gamma+\delta)=38/3$$ Let $$\alpha+\beta=k_1 $$ $$ \gamma+\delta=k_2$$ $$\implies k_1+k_2=25/3$$ & $$k_1 k_2=38/3$$ Solving first we get $$k_1 ,k_2$$ Then we get the roots as $$6,1/3,1+i,1-i$$ Now the question that arises is that can we generally solve the quartic by factorization if only a single information is given about its roots and any one of the coefficients is unknown?Also will the way I solved this question be always applicable or are there certain restrictions?

P.S: An explanation with elementary algebraic concepts will be preferred.

  • $\begingroup$ I do not have an idea what we can do if the quadratic coefficient is unknown. $\endgroup$ – Peter Apr 5 '16 at 17:15
  • $\begingroup$ @Peter For the rest of the coefficients are you sure it can be solved with a single information about its roots? $\endgroup$ – katipra Apr 5 '16 at 17:18
  • $\begingroup$ If we know the product of two roots and only the coefficient $a_3$ or $a_1$ is missing, you have shown very elegant that we can find out the missing coefficient and solve the equation. I am not sure yet about the cases $a_4$ and $a_0$ being the missing coefficient. $\endgroup$ – Peter Apr 5 '16 at 17:20
  • $\begingroup$ @Peter If the information about the roots be arbitrary like say sum of any two roots or any algebraic function of the roots, can you still determine the missing $a_1 $or $ a_3$ ? $\endgroup$ – katipra Apr 5 '16 at 17:37
  • $\begingroup$ Probably not. At least under the conditions I mentioned your solution always works. $\endgroup$ – Peter Apr 5 '16 at 17:42

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