Chapter 8.7, Problem 10MRE

To determine

To solve the equation.

Answer to Problem 10MRE

The values of $y\text{ is }-2\text{,3,}\left(1\pm i\right)$

Explanation of Solution

Given information:

  $y^4-3y^3-2y^2+10y-12=0;1+i$

Calculation:

The equation can be obtained as

  $y^4-3y^3-2y^2+10y-12=0;1+i$

  $\left(1+i\right)$ is a root, this means $\left(1-i\right)$ is also a root since coefficient are real

  $\begin{array}{l}y=\left(1-i\right)\text{ or }y=\left(1+i\right)\\ \left[y-\left(1-i\right)\right]=0\text{ or }\left[y-\left(1+i\right)\right]=0\\ \left(y-1+i\right)=0\text{ or }\left(y-1-i\right)=0\\ \text{multiply both the equation}\\ \left(y-1+i\right)\left(y-1-i\right)\\ y\left(y-1-i\right)-1\left(y-1-i\right)+i\left(y-1-i\right)\\ y^2-y-yi-y+1+i+yi-i-i^2\\ y^2-2y+1+1\left[\text{since,}{i}^2=-1\right]\\ y^2-2y+2\end{array}$

Divide the equation $y^4-3y^3-2y^2+10y-12$ by $y^2-2y+2$

  $\begin{array}{l}{y}^2-2y+2\begin{array}{c}\hfill y^2-y-6\\ \hfill \overline{)y^4-3y^3-2y^2+10y-12}\end{array}\\ \underset{\_}{y^4-2y^3+2y^2}\\ -y^3-4y^2+10y\\ -\underset{\_}{y^3+2y^2-2y}\\ -6y^2+12y-12\\ \underset{\_}{-6y^2+12y-12}\\ \text{ 0}\end{array}$

Therefore,

  $\begin{array}{l}\left(y^2-y-6\right)\left(y^2-2y+2\right)\\ =[y^2-\left(3y-2y\right)-6]\left(y^2-2y+2\right)\left[\text{write }y=\left(3y-2y\right)\right]\\ =\left(y^2-3y+2y-6\right)\left(y^2-2y+2\right)\\ =\left[y\left(y-3\right)+2\left(y-3\right)\right]\left(y^2-2y+2\right)\\ =\left(y-3\right)\left(y+2\right)\left(y^2-2y+2\right)\end{array}$

Apply Sridhar Acharya formula to solve the equation $\left(y^2-2y+2\right)$

  $\begin{array}{l}\left(y^2-2y+2\right)\\ y=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(1\right)\left(2\right)}}{2\left(1\right)}\\ a=\frac{2\pm \sqrt{4-8}}{2}\\ a=\frac{2\pm \sqrt{-4}}{2}\\ c=\frac{2\pm 2i}{2}\\ c=1\pm i\left[\text{divide numerator and denominator by 2}\right]\end{array}$

Therefore,

  $\begin{array}{l}\left(y-3\right)=0\text{ or }\left(y+2\right)=0\text{ or }y=1\pm i\\ y=3\text{ or }y=-2\end{array}$

Hence,

The values of $y\text{ is }-2\text{,3,}\left(1\pm i\right)$