Question

The product of two of the four roots of the quadratic equation $x^4-{18}^3+k{x}^2+200x-1984=0$ is $-32.$ Determine the value of $k.$
(correct answer + 3, wrong answer 0)

• A. 86
• B. 86.00
• C. 86.0

Answer 1

Answer: (A), (B), (C)

Let $a,b,c,d$ be the four roots of $x^4-{18}^3+k{x}^2+200x-1984=0$ such that $ab=-32.$
Then, $a+b+c+d=18$
$ab+bc+cd+da+ac+bd=k$
$abc+abd+acd+bcd=-200$
$abcd=-1984$
Since $ab=-32$ and $abcd=-1984,$
we have, $cd=62$
Then, from $abc+abd+acd+bcd=-200,$ we have
$-200=-32c-32d+62a+62b=-32(c+d)+62(a+b)$
Solving this equation together with the equation $a+b+c+d=18$ gives
$a+b=4,$ $c+d=14$

From $ab+bc+cd+da+ac+bd=k$, we have
$k=ab+bc+cd+da+ac+bd=-32+ac+ad+bc+bd+62$
$\Rightarrow k=30+(a+b)(c+d)=30+4\times 14=86$
$\Rightarrow k=86$