Chapter 7.2, Problem 3WE

To determine

Solve the equation $y^2-4y+13=0$ .

Answer to Problem 3WE

The solution set for the quadratic equation is $\left\{2+3i,2-3i\right\}$

Explanation of Solution

Given information: The given quadratic equation is,

  $y^2-4y+13=\mathrm{0....}\left(\text{Given}\right)$

Formula used: If $\left(\alpha ,\beta \right)$ are the roots of the quadratic equation $a{x}^2+bx+c=0$ where $\left(a\ne 0\right)$ then,

  $\alpha ,\beta =\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{....}\left(1\right)$

Calculation: If $\left(y_1,y_2\right)$ are the roots of quadratic equation $y^2-4y+13=0$ then with the help of (1) $\left(y_1,y_2\right)$ are evaluated as,

  $\begin{array}{c}{y}_1,y_2=\frac{-\left(-4\right)\pm \sqrt{{\left(-4\right)}^2-4\left(1\right)\left(13\right)}}{2\left(1\right)}\\ =\frac{4\pm \sqrt{16-52}}{2}\\ =\frac{4\pm \sqrt{-36}}{2}\\ =\frac{4\pm 6i}{2}\mathrm{....}\left(i=\sqrt{-1}\right)\\ =2\pm 3i\end{array}$