Chapter 6.8, Problem 46WE

To determine

To show by substitution that $1-3i$ is the solution of given equation.

Explanation of Solution

Given information:

Given equation is $x^2-2x+10=0$

To prove this we need to substitute the value of $x=1-3i$ in the equation $x^2-2x+10=0$

Now we know that the value of $i^2=-1$

After putting $x=1-3i$ we shall solve the left hand side of the equation.

  ${\left(1-3i\right)}^2-2\left(1-3i\right)+10$

Here we shall use identity ${\left(a-b\right)}^2=a^2+b^2-2ab$

  $\begin{array}{l}\Rightarrow \left(1+9i^2-6i\right)-2+6i+10\\ \Rightarrow 1-9-6i-2+6i+10\\ \Rightarrow 11-11+6i-6i=0\end{array}$

As we confirmed that left hand side of equation is equal to right hand side. Hence $1-3i$ is the solution of the given equation.