How would I find the real root, I know I can say another root is (5+i) but would I use the product of the roots at all?
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4 Answers
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Complex roots always come in pairs via the conjugate, so you also know that $5-i$ is a root. Hence, you can factor your cubic to something like $$(x-(5+i))(x-(5-i))(x-A)$$ where $A$ is your yet-to-be determined real root. Expand the quantity $(x-(5+i))(x-(5-i))(x-A)$ and set it equal to $x^3+px^2+6x+q$. You can pattern match coefficients and should be able to solve for $A$ without too much difficulty.
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There are two roots $5+i, 5-i$ because the coefficients are real and the roots come in complex conjugate pairs.
You can then use Vieta's Formulas, or simple expansion of $(x-a)(x-b)(x-c)$, to note that if the roots are $a=5+i, b=5-i, c$ then $ab+ac+bc=26+5c+ic+5c-ic=26+10c=6$
This approach has the benefit of isolating the coefficient whose value you know, and simplifies equations at an earlier stage.
answered Mar 23, 2015 at 18:27
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Sum of roots = $ - p = 5 + i + 5 - i + U $; Unknown real root U = $ -p -10. $
answered Mar 23, 2015 at 18:53
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$\begingroup$ This was my answer too, but then I realised that they want the actual value of the root (not just in terms of $p$ and $q$). $\endgroup$– TonyKMar 24, 2015 at 9:12
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Hints:
The following quadratic is a factor of your polynomial:
$$\begin{cases}&(5-i)(5+i)=|5-i|^2=26\\{}\\&5-i+5+i=10\end{cases}\;\;\implies (x-(5-i))(x-(5+i))=x^2-10x+26$$
