Question

Solve the following quadratic equation:
(iv) $x^2-(2+i)x-(1-7i)=0$

Answer 1

We have, $x^2-(2+i)x-(1-7i)=0$
On comparing with general form of quadratic equation
$a{x}^2+bx+c=0$
We get $a=1,b=-(2+i),c=-(1-7i)$
Roots of the equation are

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{(2+i)\pm \sqrt{{(-(2+i\left)\right)}^2-4\times 1\times \{-(1-7i\left)\right\}}}{2\times 1}$

$=\frac{(2+i)\pm \sqrt{(2^2+4i+i^2)+(4-28i)}}{2}$

$=\frac{(2+i)\pm \sqrt{(4+4i-1)+(4-28i)}}{2}$

$x=\frac{(2+i)\pm \sqrt{7-24i}}{2}$

$=\frac{(2+i)\pm \sqrt{16-24i-9}}{2}$

$=\frac{(2+i)\pm \sqrt{4^2-2\times 4\times 3i+{\left(3i\right)}^2}}{2}$

$=\frac{(2+i)\pm \sqrt{{(4-3i)}^2}}{2}=\frac{(2+i)\pm (4-3i)}{2}$

$=\frac{(2+i)+(4-3i)}{2},\frac{(2+i)-(4-3i)}{2}$

$=\frac{6-2i}{2},\frac{-2+4i}{2}$

$=(3-i),(-1+2i)$

Hence, the roots are $(3-i)and(-1+2i)$