Question

Solve the given quadratic equation:
x² - (5 - i)x + (18 + i) = 0

Answer 1

Answer:

The roots of the given equation are $\dfrac{(5-i)+ \sqrt{-48-14i}}{2}$ and $=\dfrac{(5-i)- \sqrt{-48-14i}}{2}$.

Step-by-step explanation:

Given quadratic equation is

• x² - (5 - i)x + (18 + i) = 0.

where,

$\rm i=\sqrt{-1}$.

The root of a quadratic equation of the form

$\rm ax^2+bx+c=0$

are given by

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

Now, comparing the given quadratic equation with this equation,

• $\rm a=1$.
• $\rm b=-(5-i)$.
• $\rm c=(18+i)$.

The roots of the given equation is given by

$\rm x=\dfrac{-(-(5-i))\pm \sqrt{(-(5-i))^2-4\cdot(1)\cdot (18+i)}}{2}\\=\dfrac{(5-i)\pm \sqrt{(5^2+i^2-10i)-(72+4i)}}{2}\\=\dfrac{(5-i)\pm \sqrt{(25-1-10i)-(72+4i)}}{2}\ \ \ \ \ \ \because (i^2=-1)\\=\dfrac{(5-i)\pm \sqrt{(25-1-72)-4i-10i}}{2}\\ =\dfrac{(5-i)\pm \sqrt{-48-14i}}{2}.$