One of the roots of the equation \(x^3+2x^2+ax+b=0\), where \(a\) and \(b\) are real, is \(1+\frac12\mathrm{i}\). Find the other roots of the equation and the values of \(a\) and \(b\).[5]
\(1-\frac12\mathrm{i}\) is also a root (real coefficients).
\(r_3=-2-\left(1+\frac12\mathrm{i}\right)-\left(1-\frac12\mathrm{i}\right)=-4\)
\((x+4)\left((x-1)^2+\frac14\right)=x^3+2x^2-\frac{27}{4}x+5\)
\(\therefore a=-\frac{27}{4},\quad b=5\)
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