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- Feb 14, 2012

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- Thread starter
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- #1

- Feb 14, 2012

- 3,756

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Feb 14, 2012

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Congratulations to the following members for their correct solutions:

1. MarkFL

2. Ackbach

3. Pranav

Solutions from Ackbach and Pranav are explained in detail below:

Solution from Ackbach:

$$x^3+kx-2(k+4)=(x-a)(x-b)^2=(x-a)(x^2-2bx+b^2)=x^3-ax^2-2bx^2+2abx+b^2x-ab^2$$

$$=x^3-(a+2b)x^2+(b^2+2ab)x-ab^2.$$

This sets up a system of three equations in three unknowns, by equating coefficients of like powers:

\begin{align*}

a+2b&=0 \\

b^2+2ab&=k \\

-ab^2&=-2(k+4).

\end{align*}

From the first equation, we have $a=-2b$. Plugging this into the second two equations yields

\begin{align*}

-3b^2&=k \\

2b^3&=-2(k+4).

\end{align*}

Next, we have

$$2b^3=-2(-3b^2+4) \implies b^3-3b^2+4=0.$$

This cubic has a solution $b=2$; dividing out the factor $b-2$ yields the depressed quadratic $b^2-b-2=(b+1)(b-2)$. Hence, the distinct solutions are $b=-1, 2$. The corresponding $k$ values are $k=-3, -12$.

Solution from Pranav:

$$\begin{align*}

x^3+kx-2(k+4) =& (x-a)^2(x-b) \\ =& x^3-(2a+b)x^2+(2ab+a^2)x-a^2b\\

\end{align*}$$

Comparing the coefficient of $x^2$ on both sides gives $0=2a+b \Rightarrow b=-2a$.

Comparing the coefficient of $x$ on both sides gives $k=2ab+a^2 \Rightarrow k=-3a^2 \,\,\,(\because b=-2a)$ .

Comparing the constant terms gives $2(k+4)=a^2b \Rightarrow 2(-3a^2+4)=-2a^3 \Rightarrow a^3-3a^2+4=0$.

The above cubic has $a=-1$ as one of it roots. Dividing the cubic by $a+1$ gives $a^2-4a+4=0 \Rightarrow (a-2)^2=0 \Rightarrow a=2$.

The values of $k$ can be found using $k=-3a^2$. Hence the values of $k$ are $-3$ and $-12$.

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