Chapter 1.8, Problem 65E

Solution

To write: The quadratic equation in the form $a{x}^2+bx+c=0$ where $c=4$ and the equation has the solutions $-1+i\text{ and }-1-i$ .

The required quadratic equation is $x^2+2x+2=0$ .

Given:

The quadratic equation is $a{x}^2+bx+c=0$ where $c=4$ . So, the quadratic equation can be written as:

  $a{x}^2+bx+4=0\mathrm{......}\left(1\right)$

Concept used:

If $\left\{\alpha ,\beta \right\}$ are the solutions of the quadratic equation $a{x}^2+bx+c=0$ where $a\ne 0$ , then the solutions $\left\{\alpha ,\beta \right\}$ must satisfy $a{x}^2+bx+c=0$ .

Calculation:

Since, $-1+i\text{ and }-1-i$ are the solutions of the quadratic equation. Therefore, they will satisfy $\left(1\right)$ ,

  $\begin{array}{c}a{\left(-1+i\right)}^2+b\left(-1+i\right)+4=0\mathrm{......}\left(\text{Satisfy }-1+i\right)\\ a\left(1+i^2-2i\right)+b\left(-1+i\right)+4=0\\ -b+i\left(b-2a\right)+4=0\\ b-i\left(b-2a\right)=4\mathrm{......}\left(2\right)\\ a{\left(-1-i\right)}^2+b\left(-1-i\right)+4=0\mathrm{......}\left(\text{Satisfy }-1-i\right)\\ a\left(1+i^2+2i\right)-b\left(1+i\right)+4=0\\ -b+\left(2a-b\right)i+4=0\\ \text{or, }b-\left(2a-b\right)i=4\mathrm{......}\left(3\right)\end{array}$

Subtract $\left(3\right)$ from $\left(2\right)$ gives,

  $\begin{array}{c}b-i\left(b-2a\right)=b-\left(2a-b\right)i\\ b-2a=2a-b\\ 4a=2b\\ b=2a\mathrm{......}\left(4\right)\\ b-i\left(b-2a\right)=4\mathrm{......}\left(\text{From }\left(2\right)\right)\\ 2a-i\left(2a-2a\right)=4\mathrm{......}\left(\because b=2a\right)\\ 2a=4\\ a=2\end{array}$

Substitute $a=2$ in $\left(4\right)$ gives $b=4$ .

Now substitute, $a=2$ and $b=4$ in $\left(1\right)$ gives,

  $\begin{array}{c}\left(2\right)x^2+\left(4\right)x+4=0\\ 2x^2+4x+4=0\\ x^2+2x+2=0\end{array}$

Hence, the required quadratic equation is $x^2+2x+2=0$ .