Let $x_1,x_2,x_3,x_4$ denote the four roots of the equation $$x^4 + kx^2 + 90x – 2009 = 0.$$ If $$x_1x_2 = 49,$$ then find the value of $k$.
What I tried :- From Vieta's Formula for quartic equations I get that
$$ x_1 + x_2 + x_3 + x_4 = 0$$
$$x_1x_2 + x_1x_3 + x_2x_3 + x_2x_4 + x_3x_4 + x_4x_1 = k$$
$$x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_1 + x_4x_1x_2 = (-90)$$
$$ x_1x_2x_3x_4 = -2009$$
Since $x_1x_2 = 49$, we have $x_3x_4 = (-41)$.
Hence in equation $3$ I get
$$x_1x_2(x_3 + x_4) + x_3x_4(x_1 + x_2) = (-90)$$
$$\Rightarrow 49(x_3 + x_4) – 41(x_1 + x_2) = (-90)$$
As $(x_1 + x_2 + x_3 + x_4) = 0$, $(x_1 + x_2) = -(x_3 + x_4)$, so $$49(-x_1 – x_2) – 41(x_1 + x_2) = (-90)$$
$$\Rightarrow (-90)(x_1 + x_2) = (-90)$$
$$\Rightarrow (x_1 + x_2) = 1,$$ so $(x_3 + x_4) = (-1)$.
From here I don't know how to proceed, can anyone help?
Best Answer
Hint:
$$(x-x_1)(x-49/x_1)(x-x_3)(x+41/x_3)$$
$$=\left(x^2-x\left(x_1+\dfrac{49}{x_1}\right)+49\right)\left(x^2-x\left(x_3-\dfrac{41}{x_3}\right)-41\right)$$
$x_1+\dfrac{49}{x_1}+x_3-\dfrac{41}{x_3}=0\implies x_1+x_3=\dfrac{41x_1-49x_3}{2009}$
This will give $\dfrac{x_3}{x_1}=?$
$$-90=x_1x_2(x_3+x_4)+(x_1+x_2)x_3x_4=49(x_3+x_4)-41(x_1+x_2)$$
Replace the values of $x_3$ in terms of $x_1, x_2=\dfrac{49}{x_1}, x_4=-\dfrac{41}{x_3}=\dfrac ?{x_1}$