Question

Solve the following quadratic equation:
(iii) $(2+i)x^2-(5-i)x+2(1-i)=0$

Answer 1

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

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

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

$=\frac{(5-i)\pm \sqrt{(25-10i+i^2)-8(2+i)(1-i)}}{2(2+i)}$

$x=\frac{(5-i)\pm \sqrt{(25-10i-1)-8(2-2i+i=i^2)}}{2(2+i)}$

$=\frac{(5-i)\pm \sqrt{(24-10i)-8(2-i+1)}}{2(2+i)}$

$=\frac{(5-i)\pm \sqrt{(24-10i)-(24-8i)}}{2(2+i)}$

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

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

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

$x=\frac{(5-i)\pm \sqrt{{(1-i)}^2}}{2(2+4)}=\frac{(5-i)\pm (1-i)}{2(2+i)}$

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

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

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

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

$x=\frac{(6-2i-3i+i^2)}{2^2-i^2},\frac{4-2i}{2^2-i^2}$

$=\frac{(6-5i-1)}{4+1},\frac{4-2i}{4+1}$

$=(1-i),\frac{4-2i}{5}$

Hence, the roots are $(1-i)\&(\frac{4}{5}-\frac{2i}{5})$