#### Question

The product of two of the four roots of the quartic equation x^4 − 18x^3 + kx^2 + 200x − 1984 = 0 is −32. Determine the value of k.

Collected in the board: Quartic Equations

Steven Zheng posted 1 hour ago

Let a,b,c,d are the four roots of the quartic equation x^4 − 18x^3 + kx^2 + 200x − 1984 = 0

The following equations are obtained by using the Vieta's formula for a quartic equation.

a+b+c+d = 18
(1)
(2)
abc+acd+bcd+abd = -200
(3)
abcd = -1984
(4)

Since the product of two of the four roots is -32, let ab = -32, then cd = 62

Substitute ab and cd to the equation (3)

-32c+62a+62b-32d = -200

Cancel the common factor

31a+31b-16c-16d = -100
(5)

Subtract (5) from 31*(1)

47(c+d) = 658

Then

c+d = \dfrac{658}{47}
(6)

Similarly, 16*(1)+(5) gives

47(a+b) =188

a+b = \dfrac{224}{47}
(7)

Square (6) and subtract 2cd gives the value of

c^2+d^2=\Big( \dfrac{658}{47}\Big)^2-124

Square (7) and subtract 2ab gives the value of

a^2+b^2 = \Big(\dfrac{188}{47} \Big)^2+64

Then,

a^2+b^2+c^2+d^2 =152

Therefore,

= \dfrac{1}{2} [(a+b+c+d)^2 - (a^2+b^2+c^2+d^2)]

=\dfrac{1}{2}(18^2- 152 )

=86

Verification

Solving the quartic equation x^4 − 18x^3 + 86x^2 + 200x − 1984 = 0 gives four roots

\begin{cases} x_1 = -4 \\ x_2 = 8 \\ x_3 = 7+\sqrt{13}i \\ x_4 = 7-\sqrt{13}i \end{cases}

the product of the two roots is indeed equal to -32 and the product of the four equals to -1984

Steven Zheng posted 1 hour ago

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