Question

Solve the given quadratic equation:
x² - (2 + i)x - (1 - 7i) = 0

Answer 1

Solution:

As we know that Quadratic equation can be easily Solve by Quadratic formula

Standard Quadratic equation
$a {x}^{2} + bx + c = 0 \\ \\ x_{1,2} = \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a}$
here equation is

${x}^{2} - (2 + i)x - (1 - 7i) = 0 \\ \\ x_{1,2} = \frac{(2 + i) ±\sqrt{ {(2 + i)}^{2} + 4(1 - 7i) } }{2} \\ \\ x_{1,2} = \frac{(2 + i) ± \sqrt{4 - 1 + 4i + 4 - 28i} }{2} \\ \\ x_{1,2}= \frac{(2 + i) ± \sqrt{7 - 24i} }{2}$
So,
$x_{1} = \frac{(2 + i) + \sqrt{7 - 24i} }{2} \\ \\ x_{2} = \frac{(2 + i) - \sqrt{7 - 24i} }{2}$
Hope it helps you.

Answer 2

Answer:

3 - i and - 1+2i

Step-by-step explanation:

Concept:

The roots of the quadratic equation can be found by using the quadratic formula

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

x² - (2 + i)x - (1 - 7i) = 0

Now,

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

$x=\frac{2+i+\sqrt{(2+i)^2+4(1-7i)}}{2}\:and\:x=\frac{2+i-\sqrt{(2+i)^2+4(1-7i)}}{2}\\\\x=\frac{2+i+\sqrt{4-1+4i+4-28i}}{2}\:and\:x=\frac{2+i-\sqrt{4-1+4i+4-28i}}{2}$

$x=\frac{2+i+\sqrt{7-24i}}{2}\:and\:x=\frac{2+i-\sqrt{7-24i}}{2}$

$x=\frac{2+i+4-3i}{2}\:and\:x=\frac{2+i-(4-3i)}{2}$

$x=\frac{2+i+4-3i}{2}\:and\:x=\frac{2+i-4+3i}{2}$

$x=\frac{6-2i}{2}\:and\:x=\frac{-2+4i}{2}$

$x=\frac{2(3-i)}{2}\:and\:x=\frac{2(-1+2i)}{2}$

$x=3-i\:and\:x=-1+2i$

Note:

7-24i = 4²+(3i)² - 2(4)(3i)

7-24i = (4 -3i) ²

√7-24i = 4 -3i