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# Assignment 16, 2iv

I don't get how I'm supposed to do it. It tells me to use substitution of x = ky to find the second equation, though I've done this and the k values fluctuate. Moreover, what are we supposed to do about the constant at the end without any x attached to it?

### 2 iv

Substituting $x=ky$ into $2x^3-5x^2-6x+9=0$ gives:
\begin{align*}
2(ky)^3 - 5(ky)^2-6(ky)+9&=0\\
2k^3y^3-5k^2y^2-6ky+9&=0
\end{align*}
You really want this to be a multiple of $6y^3-5y^2-2y+1=0$. Looking at the constant term, it looks as if we want the equation with $k$s in to be $9 \times (6y^3-5y^2-2y+1=0)$. This means we want $5k^2y^2=9 \times 5y^2$ which suggests that $k=3$ ($k=-3$ does not work - try it!) might be a good thing to try. This gives:
\begin{align*}
2k^3y^3-5k^2y^2-6ky+9&=0\\
2 \times 27 y^3 - 5 \times 9 y^2 - 6 \times 3y + 9&=0\\
2 \times 3y^3 - 5y^2 - 2y + 1&=0\\
6y^3-5y^2-2y+1&=0
\end{align*}
Therefore if we make the substitution $x=3y$ the equation $2x^3-5x^2-6x+9=0$ becomes $6y^3-5y^2-2y+1=0$. The solutions to the second equation can be found by using $y=\frac 1 3 x$ where the $x$ values are the solutions to the first equation.

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