Cubic Question 1

L3CALC

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Given the cubic function

$$2x^3+px-(3+5i)x+q=0$$

with roots $i,\frac{1}{k},-(1+k)$,

Find the value of $p$, $q$, $k$. But $k\in \mathbb{R}$.

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$\boxed{正}\text{: Comparison of Coefficients}$

$$2x^3+px-(3+5i)x+q=0$$ $$\equiv 2(x-i)\left(x-\frac{1}{k}\right)(x+(1+k))=0$$ $$⟹2\left[\left(x^2-\frac{x}{k}-ix+\frac{i}{k}\right)(x+(k+1))\right]=0$$ $$⟹2\left[x^3+\left(k+1-\frac{1}{k}-i\right)x^2+\left(-\frac{k+1}{k}-i(k+1)+\frac{i}{k}\right)x+\frac{k+1}{k}i\right]=0$$ $$⟹2x^3+2\left(k+1-\frac{1}{k}-i\right)x^2+2\left(-\frac{k+1}{k}-i(k+1)+\frac{i}{k}\right)x+\frac{2(k+1)}{k}i=0$$

Now comparing coefficients:

$$p=2\left(k+1-\frac{1}{k}-i\right)$$ $$-3-5i=2\left(-\frac{k+1}{k}-i(k+1)+\frac{i}{k}\right)$$ $$q=\frac{2(k+1)}{k}i$$

Using the second equation and by equating $\mathrm{Re}=\mathrm{Re}$, $\mathrm{Im}=\mathrm{Im}$:

$$-3=-\frac{2(k+1)}{k}$$ $$3k=2k+2$$ $$\boxed{k=2}$$

Now substitute into other equations:

$$\boxed{p=5-2i}$$ $$\boxed{q=3i}$$

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$\boxed{\text{AS}}:\text{Vieta’s Formulas}$ Vietas formulas of a cubic function states that for a cubic function

$$a{x}^3+b{x}^2+cx+d=0$$

with roots $\alpha ,\beta ,\gamma$, the following relationship can be formed:

$$\alpha +\beta +\gamma =-\frac{b}{a}$$ $$\alpha \beta +\alpha \gamma +\beta \gamma =\frac{c}{a}$$ $$\alpha \beta \gamma =-\frac{d}{a}$$

Using this, we can use the relationship between the roots and the coefficients of the cubic:

$$i+\frac{1}{k}-(k+1)=-\frac{p}{2}$$ $$\frac{-3-5i}{2}=\frac{i}{k}-i(k+1)-\frac{k+1}{k}$$ $$-\frac{q}{2}=\frac{i(k+1)}{k}$$

Using the second equation and by equating $\mathrm{Re}=\mathrm{Re}$, $\mathrm{Im}=\mathrm{Im}$:

$$\mathrm{Re}:k=2$$ $$\mathrm{Im}:k=2,k=-\frac{1}{2}$$

but $-\frac{1}{2}$ subbed back to $\mathrm{Re}$ is false, so reject, and therefore $\boxed{k=2}$. Substituting $k=2$ into first and third equations:

$$\boxed{p=5-2i}$$ $$\boxed{q=3i}$$