Question

Solve $x^2-(7-i)x+(18-i)=0$ .

Answer 1

$x^2-(7-i)x+(18-i)=0$

equation is in the form $a{x}^2+bx+c=0$

$a=1,b=-(7-i),c=18-i$

$\alpha =\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\beta =\frac{-b-\sqrt{b^2-2ac}}{2a}$

$\alpha =\frac{(7-i)+\sqrt{(7-i{)}^2-4(18-i)}}{2}$ $\beta =\frac{(7-i)-\sqrt{(7-i{)}^2-4(18-i)}}{2}$

$\alpha =\frac{(7-i)\sqrt{-24-10i}}{2}$ $\beta =\frac{(7-i)-\sqrt{-24-10i}}{2}$

Now for the value of $\sqrt{24-10i}$

Let $a+ib=\sqrt{-24-10i}$

$a^2-b^2+2abi=-24-10i$

$a^2-b^2=-242ab=-10$

Consider

$(a^2+b^2{)}^2=(a^2-b^2{)}^2+4a^2{b}^2$

$=(-24{)}^2+(-10{)}^2=6+6$

$a^2+b^2=26$

Now,

$a^2-b^2=-24$ and $a^2+b^2=26$

$a^2=1b^2=25$

$a=\pm 1b=\pm 5$

$2ab=-10$ So, $a$ and $b$ must be opposite signs

$\therefore a=1$ and $b=-5$ or $a=-1$ and $b=5$

$\Rightarrow \sqrt{-24-100}=1-5i$ or $-1+5i$

Let us consider $1-5i$

$\alpha =\frac{7-i+(1-5i)}{2}$ and $\beta =\frac{(7-i)-(1-5i)}{2}$

$\alpha =4-3i\beta =3-3i$

$\therefore$ Roots are $4-3i$ and $3-3i$