## - Roots of Equations

In general, if a polynomial, with real coefficients, has complex roots, they occur in conjugate pairs. For example, if $$2+i$$ is the root of a polynomial equation then $$2-i$$ is also a root of the same equation.

RECALL: If $$\displaystyle x^2+{b\over a} x+{c\over a}=0$$, then $$\displaystyle α+β=-{b\over a}$$ and $$\displaystyle αβ={c\over a}$$ where $$α$$ and $$β$$ are the roots of the equation i.e.
$$x^2-$$(sum of roots)$$x+$$(product of roots)$$=0$$
Also, if $$\displaystyle x^3+{b\over a} x^2+{c\over a} x+{d\over a}=0$$ then $$\displaystyle α+β+γ=-{b\over a}$$, $$\displaystyle αβ+βγ+αγ={c\over a}$$ and $$\displaystyle αβγ=-{d\over a}$$ where $$α, β$$ and $$γ$$ are the roots of the equation.

LESSON 1
Given one root find the equation
(i) $$5i$$
(ii) $$4-3i$$
SOLUTION
(i) Let $$α=5i$$ then $$β=-5i$$
$$α+β=(5i)+(-5i)=0$$
$$αβ=(5i)(-5i)=25$$
Equation is $$x^2+25=0$$

(ii) Let $$α=4-3i$$ then $$β=4+3i$$
$$α+β=(4-3i)+(4+3i)=8$$
$$αβ=(4-3i)(4+3i)=25$$
Equation is $$x^2-8x+25=0$$

LESSON 2
Given that $$1-2i$$ is a root of the equation $$x^3+x^2-x+15=0$$, find the other 2 roots.
SOLUTION
Since complex roots occur in conjugate pairs and a cubic polynomial has 3 roots one root must be real.
Let $$α=1-2i, β=1+2i, γ\in\mathbb{R}$$
$$\displaystyle α+β+γ=-{b\over a}=-1$$
$$⟹(1-2i)+(1+2i)+(γ)=-1$$
$$⟹2+γ=-1$$
$$⟹γ=-3$$