Question

Two roots of a biquadratic $x^4-18x^3+k{x}^2+200x-1984=0$ have their product equal to $(-32)$. Find the value of $k$.

Answer 1

Let $4$ roots be a,b,c,d so that $ab=-32$ from here we show
$a+b+c+d=18\to 1$
$ab+ac+ad+bc+bd+cd=k\to 2$
$abc+abd+acd+bcd=-200\to 3$
$abcd=-1984\to 4$
From last $3$ equations, we see that $cd=\frac{abcd}{ab}=\frac{-1984}{-32}=62$
So second equation becomes $-32+ac+ad+bc+bd+62=k$
And so $ac+ad+bc+bd=k-30$
Let $p=a+b$ & $q=c+d$
From $3$
$-200=ab(c+d)+cd(a+b)$
$-200=-32q+62p$
We know, $a+b+c+d=18$ gives $p+q=18$. Then we have $2$ linear equation in p and q, which we solve to obtain $p=4$ & $q=14$
$\therefore$We have $\frac{(a+b)}{4}\frac{(c+d)}{14}=k-30$
$\therefore k=4\times 14+30=86$