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 $k$.

Question

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?

Answer 1

Your Vieta for $k$ is wrong and should be $k=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4$ and this can be rewritten in terms of information you already have. Give that a go.

Answer 2

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 $x_1, x_2=\dfrac{49}{x_1}, x_4=-\dfrac{41}{x_3}=\dfrac ?{x_1}$

Answer 3

$x_1+x_2+x_3+x_4=0$

$x_1x_2x_3x_4=-2009$

$x_1x_2=49$

$x_3x_4=-41$

Let $x_1=a, x_2=b, x_3=c, x_4=d$

Then $abc+bcd+acd+bda=-90$

$abd+abc+cdb+cda=-90$

$ab(d+c)+cd(a+b)=-90$

$49(d+c)-41(a+b)=-90$

And $a+b=-c-d$ , so let $a+b=m$ and $c+d=-m$

$\implies -49m-41m=-90$

$-90m=-90$

$m=1$

$\implies a+b=1,c+d=-1$

$a+b=1 \implies (a+b)^2=a^2+2ab+b^2=1$

$ab=49$

$a^2+b^2=-97$

$c+d=-1 \implies (c+d)^2=c^2+2cd+d^2=1$

$cd=-41$

$c^2+d^2=83$

$(a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+bc+cd+ad+ac+bd)$

$0=a^2+b^2+c^2+d^2+2(ab+bc+cd+ad+ac+bd)$

$-2k=a^2+b^2+c^2+d^2$

$-2k=-14 \implies k=7$

Answer 4

We have :- $x_1 + x_2 + x_3 + x_4 = 0$
$\rightarrow x_3 + x_4 = -(x_1 + x_2)$

$(x_1 * x_2 * x_3 * x_4 = -2009)$ and $(x_1 * x_2 = 49)$
$\rightarrow x_3*x_4 = -41$

Now, $$k = (x_1+x_2)(x_3+x_4) + (x_1*x_2) + x_3*x_4 = 8 + (x_1+x_2)(x_3+x_4)$$

$$-90 = x_1x_2(x_3+x_4) + x_3x_4(x_1+x_2)$$

$$\rightarrow -90 = 90(x_3+x_4)$$

$\rightarrow x_3 + x_4 = -1$ and $x_1 + x_2 = 1$.

So :- $k = (8 - 1) = 7$.