Question

Solve for x over the complex numbers.


x^2+8x+20=0




Enter your answers in the boxes in standard form.

Answer 1

Answer:

$x_1=-4+2i\\x_2=-4-2i$

Step-by-step explanation:

A quadratic function ($ax^2+bx+c=0$) usually has two solutions for x. To find that solutions we have to use Bhaskara's Formula:

$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\\and\\x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

We have the expression:

$x^2+8x+20=0$

Where,

$a=1, b=8,c=20$

Then, applying Bhaskara's Formula:

$x_1=\frac{-8+\sqrt{8^2-4.1.20}}{2.1}=\frac{-8+\sqrt{64-80}}{2}\\\\x_1=\frac{-8+\sqrt{-16}}{2}=\frac{-8+\sqrt{-1}\sqrt{16}}{2}\\\\x_1=\frac{-8+i4}{2}\\\\x_1=-4+2i$

Remember that: $\sqrt{-1}=i$

We have the first solution for x, now we have to find the second:

$x_2=\frac{-8-\sqrt{8^2-4.1.20}}{2.1}=\frac{-8-\sqrt{64-80}}{2}\\\\x_2=\frac{-8-\sqrt{-16}}{2}=\frac{-8-\sqrt{-1}\sqrt{16}}{2}\\\\x_2=\frac{-8-i4}{2}\\\\x_2=-4-2i$

Then the solutions for x are:

$x_1=-4+2i\\x_2=-4-2i$

Over the complex numbers.