- 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 \(α=1-2i, β=1+2i, γ\in\mathbb{R}\)
\(\displaystyle α+β+γ=-{b\over a}=-1\)
\(⟹(1-2i)+(1+2i)+(γ)=-1\)
\(⟹2+γ=-1\)
\(⟹γ=-3\)