Chapter MPS, Problem 6.11MPS

To determine

To Find: The expression from the given roots.

Answer to Problem 6.11MPS

  $x^2+\left(3i-5\right)x-\left(6-3i\right)=0$

Explanation of Solution

Given:

  $Zeros=-1\&6-3i$

Formula Used:

  $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$

Calculation:

  $\begin{array}{l}Zeros=-1\&6-3i\\ \left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\\ a=-1\\ b=6-3i\\ \text{⇒}\left(x+1\right)\left(x-6+3i\right)\\ \text{⇒}{x}^2-6x+3xi+x-6+3i\\ \text{⇒}{x}^2+\left(3i-5\right)x-\left(6-3i\right)=0\end{array}$

Conclusion:

Thus the expression is $x^2+\left(3i-5\right)x-\left(6-3i\right)=0$ .