The equation ${{z}^{4}}-4{{z}^{3}}+a{{z}^{2}}-20z+25=0,\mathrm{ }a\in \mathbb{R},$ has a root \(k\

Question

The equation ${{z}^{4}}-4{{z}^{3}}+a{{z}^{2}}-20z+25=0,\mathrm{ }a\in \mathbb{R},$ has a root $k\mathbf{i}$, where $k$ is a real number. Explain clearly why there is more than one possible value for $k$.[1] Find the possible exact values of $k$ and show that $a=10$. Hence find the roots of the above equation.[7] Deduce the roots of the following equations in the form $x+\mathbf{i}y,$ where $x$ and $y$ are real, ${{w}^{4}}+4\mathrm{i}{{w}^{3}}-10{{w}^{2}}-20\mathrm{i}w+25=0,$ $25{{v}^{4}}-20{{v}^{3}}+10{{v}^{2}}-4v+1=0.$[4]

Accepted Answer

$\text{(i)}$ Since the coefficients are real, both $k\mathbf{i}$ and $-k\mathbf{i}$ are roots $\text{(ii)}\ k=\pm\sqrt5,\ a=10$, with roots $z=\pm\sqrt5\mathbf{i},\ 2\pm\mathbf{i}$ $\text{(iii)(a)}\ w=\pm\sqrt5,\ 1-2\mathbf{i},\ -1-2\mathbf{i}$ $\text{(iii)(b)}\ v=\pm\dfrac{\mathbf{i}}{\sqrt5},\ \dfrac{2+\mathbf{i}}5,\ \dfrac{2-\mathbf{i}}5$

2013 DHS P2 Q4

2013 DHS P2 Q4

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