Determine k and solve the equation x^3 -kx^2 + 20x +12 = 0.

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- Aug 30th 2007, 06:43 PMMr_Greensolving an equation
Determine k and solve the equation x^3 -kx^2 + 20x +12 = 0.

Thanks - Aug 30th 2007, 06:45 PMThePerfectHacker
- Aug 30th 2007, 06:47 PMMr_Green
my bad...

Determine k and solve the equation x^3 - kx^2 +20x + 12 = 0, if one of its zeros is the triple of another. - Aug 30th 2007, 07:03 PMThePerfectHacker
Let $\displaystyle a,b,c$ be zero's (counting multiplicity).

One of its zeros, say $\displaystyle c$, is triple of another, say $\displaystyle a$. Thus, $\displaystyle c=3a$.

Thus, we have that,

$\displaystyle \left\{ \begin{array}{c}a+b+3a = k\\ab+3a^2+3ab=20\\ 3a^2b= -12 \end{array}\right.$

Thus,

$\displaystyle \left\{ \begin{array}{c}4a+b = k\\ 4ab+3a^2 = 20 \\ a^2b = -4 \end{array}\right.$

Now we solve for $\displaystyle a,b,k$ ...

But, I have an easier way. Solving for that equation can get messy. Instead, let us*hope*that its solutions are integers. Not necessarily this approach might work but let us try it, if it works then we can save a lot of work. If it does not we can try another way.

Note,

$\displaystyle a^2b=-4$.

Now, $\displaystyle a^2>0$ so $\displaystyle b<0$. It must be that $\displaystyle a^2 \mbox{ and }b$. Are**factors**of $\displaystyle 4$. Since $\displaystyle a^2$ must be a square and positive. It means $\displaystyle a^2 = 4 \mbox{ or }1$. Thus, the solutions are $\displaystyle (a^2,b) = (4,-1) \mbox{ or } (1,-4)$. Now take the square root to remove the square on $\displaystyle a$ to get: $\displaystyle (a,b) = (-2,-1), (2,-1), (-1,-4) , (1,-4)$.

Now which of these pairs satisfies equation 2, i.e. $\displaystyle 4ab+3a^2 = 20$.

We easily see that $\displaystyle (-2,-1)$ is the only solution.

Thus, $\displaystyle a=-2,b=-1,c=3(a)=-6$.

Thus, $\displaystyle k = -(a+b+c) = 9$.