- Solving Quadratic Equations

LESSON 1

Solve the following equations

(a) \(x^2+1=0\)
(b) \(x^2+3x+3=0\)
(c) \(4x^2-2x=-1\)
SOLUTION
We need to use the quadratic formula
\(\displaystyle x={-b±\sqrt{b^2-4ac}\over 2a}\)
to solve these equations.
(a) \(x^2+1=0\)
\(x^2=-1\)
\(x=±\sqrt{-1}\)
\(x=±i\)

(b) \(x^2+3x+3=0\)
\(\displaystyle x={-3±\sqrt{3^2-4(1)(3)}\over 2(1)}\)
\(\displaystyle x={-3±\sqrt{-3}\over 2}\)
\(\displaystyle x={-3±\sqrt{3} i\over 2}\)

(c) \(4x^2-2x=-1\)
\(4x^2-2x+1=0\)
\(\displaystyle x={-(-2)±\sqrt{(-2)^2-4(4)(1)}\over 2(4)}\)
\(\displaystyle x={2±\sqrt{-12}\over8}\)
\(\displaystyle x={2±\sqrt{4(3)(-1)}\over8}\)
\(\displaystyle x={2±2\sqrt{3}i\over 8}\)
\(\displaystyle x={1\over4}±{\sqrt{3}\over4}i\)

EQUATIONS WITH COMPLEX COEFFICIENTS
LESSON 2
Determine \(z\) such that
\(z^2+(2+2i)z-(15-10i)=0\)
SOLUTION
Again we will use the quadratic formula to solve the given equation.
\(z^2+(2+2i)z-(15-10i)=0\)
\(z^2+(2+2i)z-15+10i=0\)
\(\displaystyle z={-(2+2i)±\sqrt{(2+2i)^2-4(1)(-15+10i)}\over2(1)}\)
\(\displaystyle z={-(2+2i)±\sqrt{(2+2i)(2+2i)+60-40i}\over 2}\)
\(\displaystyle z={-(2+2i)±\sqrt{8i+60-40i}\over2}\)
\(\displaystyle z={-(2+2i)±\sqrt{60-32i}\over2}\)
  • At this stage we need to determine the square root of \(60-32i\) using the usual process.
\(\sqrt{60-32i}=a+bi\)
\(60-32i=(a+bi)^2\)
\(60-32i=a^2-b^2+2abi\)
\(a^2-b^2=60\)
\(\displaystyle ab=-16 ⟹ a=-{16\over b}\)

\(\displaystyle \left(-{16\over b}\right)^2-b^2=60\)
\(\displaystyle {256\over b^2} -b^2=60\)
\(b^4+60b^2-256=0\)
\((b^2+64)(b^2-4)=0\)
\(b^2=4\)
\(b=±2\)

When \(b=2\)
\(\displaystyle a=-{16\over2}=-8\)
\(-8+2i\)

When \(b=-2\)
\(\displaystyle a=-{16\over -2}=8\)
\(8-2i\)
  • We only need to substitute ONE of the complex square roots
\(\displaystyle z={-(2+2i)±(-8+2i)\over 2}\)
\(\displaystyle z={-(2+2i)+(-8+2i)\over 2}=-5\)
\(\displaystyle z={-(2+2i)-(-8+2i)\over2}=3-2i\)