Chapter 6.8, Problem 45WE

To determine

To show by substitution that $2+i$ is the solution of given equation.

Explanation of Solution

Given information:

Given equation is $x^2-4x+5=0$

To prove this we need to substitute the value of $x=2+i$ in the equation $x^2-4x+5=0$

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

After putting $x=2+i$ we shall solve the left hand side of the equation.

  ${\left(2+i\right)}^2-4\left(2+i\right)+5$

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

  $\begin{array}{l}\Rightarrow \left(4+i^2+4i\right)-8-4i+5\\ \Rightarrow 4-1+4i-8-4i+5\\ \Rightarrow 8-8+4i-4i=0\end{array}$

As we confirmed that left hand side of equation is equal to right hand side. Hence $2+i$ is the solution of the given equation.