Question

what is the non real complex solutions of this equation: x^2-4x+20=0

Answer 1

ANSWER

x = 2 ± 4i

EXPLANATION

We can find the solutions of the equation,

$x^2-4x+20=0$

Using the quadratic formula,

$\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}$

In this equation, a = 1, b = -4 and c = 20,

$x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4\cdot1\cdot20}}{2\cdot1}=\frac{4\pm\sqrt[]{16-80}}{2}=\frac{4\pm\sqrt[]{-64}}{2}$

Since the value under the radical is negative, there are two complex solutions. Replace the negative sign with i² and solve,

$x=\frac{4\pm\sqrt[]{i^2\cdot64}}{2}=\frac{4\pm8i}{2}=2\pm4i$

Hence, the complex solutions are 2 ± 4i