Question

Solve the following quadratic equation:
(ii) $x^2-(5-i)x+(18+i)=0$

Answer 1

We have, $x^2-(5-i)x+(18+i)=0$
On comparing with general form of quadratic equation
$a{x}^2+bx+c=0$
We get $a=1,b=-(5-i),c=(18+i)$

Roots of the equatio are

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

$=\frac{(5-i)\pm \sqrt{{(-(5-i\left)\right)}^2-4\times 1\times (18+i)}}{2\times 1}$

$=\frac{(5-i)\pm \sqrt{{(25-10i+i^2)}^2-4\times 1\times (18+i)}}{2\times 1}$

$=\frac{(5-i)\pm \sqrt{(25-10i-1)-(72+4i)}}{2}$

$=\frac{(5-i)\pm \sqrt{(-47-14i-1)}}{2}$

$=\frac{(5-i)\pm \sqrt{(-48-14i)}}{2}$

$=\frac{(5-i)\pm \sqrt{(-49-14i+1)}}{2}$

$=\frac{(5-i)\pm \sqrt{-(49+14i-1)}}{2}$

$=\frac{(5-i)\pm \sqrt{-(7^2+2\times 7\times i+i^2)}}{2}$

$=\frac{(5-i)\pm \sqrt{-{(7+i)}^2}}{2}$

$=\frac{(5-i)\pm (7+i)\sqrt{-1}}{2}$

$=\frac{(5-i)\pm (7+i)i}{2}$

$=\frac{(5-i)\pm (7i-1)}{2}=\frac{(5-i)\pm (-1+7i)}{2}$

$=\frac{(5-i)+(-1+7i)}{2},\frac{(5-i)-(-1+7i)}{2}$

$=\frac{4+6i}{2},\frac{6-8i}{2}$

$=2+3i,3-4i$

Hence, the roots are $(3-4i)and(2+3i).$