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Question :

Roots of the equation $$x^{4}+2 x^{3}-5x^{2}+7x+10=0$$ are $\alpha, \beta, \gamma, \delta$ and that of $x^{4}+a x^{3}+b x^{2}+c x+d=0$ be $\alpha+\beta+\gamma, \alpha+\beta+\delta, \alpha+\gamma+\delta ; \beta+\gamma+\delta,$ then find the value of $a+b-c-d$.

What I tried:

By applying Vieta's formula(In 2nd Equation): $$\alpha+\beta+\gamma+\alpha+\beta+\delta+\alpha+\gamma+\delta+\beta+\gamma+\delta=-a$$ $$=3(\alpha+\beta+\gamma+\delta)=-a \tag1\label{eq1} $$ and By applying in Vieta's formula in 1st equation we get: $$\alpha+\beta+\gamma+\delta=-2 \tag2\label{eq2}$$ From $\eqref{eq1}$ and $\eqref{eq2}$:

$$a=6$$ In similar way, I would find $b,c,d $ which is a very tedious and time consuming method(and it's not guaranteed that answer will come).

Is there any other way?

Hint
(Not complete answer)

Thanks

Wolgwang
  • 1,563

2 Answers2

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Idea: Perform a change of roots.

More specifically, if $x_i$ are the roots to the original equation, we want $y_i = f(x_i)$ to be the roots to the new equation. What is $f(x)$?

Hint: Like you stated, $\alpha + \beta + \gamma + \delta = -2$

$f(x) = -2 - x$

$$x^4 + ax^3 + bx^2 + cd + d = f(x) ^ 4 + 2f(x) ^3 - 5f(x)^2 + 7 f(x) + 10 .$$

Using the transformation, expanding the expression gives us

$ x^4 + 6x^3 + 7x^2 - 19x - 24 $

Calvin Lin
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So the roots of the second polynomial $p(x)$ are $-2-\alpha$, $-2-\beta$,...so if we write it in a factor form we have $$p(x)=(x+2+\alpha)(x+2+\beta)...$$ so $$p(-x-2)= (x-\alpha)(x-\beta)...=q(x)$$ where $q(x)$ is a starting polynomial. So $$p(x) = q(-x-2)=...$$

nonuser
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