Question

(2+i)x²-(5-i)x+2(1-i)=0 solve quadratic equation by formula method​

Answer 1

here,a=2+i,b=-(5-i),c=2(1-i)
formulae(d)=-b±√b square-4ac|2a
a=2+i,b=-5+i,c=2-2i
5-i±√24-10i/4+2i

Answer 2

Answer:

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

Step-by-step explanation:

Given : Quadratic equation $(2+i)x^2-(5-i)x+2(1-i)=0$

To find : Solve quadratic equation by formula method​ ?

Solution :

The quadratic formula of equation $ax^2+bx+c=0$ is given by,

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

On comparing with $(2+i)x^2-(5-i)x+2(1-i)=0$

$a=2+i,\ b=-(5-i),\ c=2(1-i)$

Substitute the values,

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

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

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

$x=\frac{5-i\pm\sqrt{24-10i-8(3-i)}}{2(2+i)}$

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

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

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