Question

Solve for x over the complex numbers.

x^2+8x+20=0




Enter your answers in the boxes in standard form.

Answer 1

x = -4 + 2i and x = -4 - 2i

Explanation:

To solve the quadratic equation x^2 + 8x + 20 = 0 over the complex numbers, you can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = 1, b = 8, and c = 20. Plug these values into the quadratic formula:

x = (-8 ± √(8^2 - 4 * 1 * 20)) / (2 * 1)

x = (-8 ± √(64 - 80)) / 2

x = (-8 ± √(-16)) / 2

Now, you have the square root of a negative number, which indicates complex solutions. You can rewrite √(-16) as 4i (where i is the imaginary unit, i.e., i^2 = -1):

x = (-8 ± 4i) / 2

Now, simplify the expression by factoring out a common factor of 4 in the numerator:

x = (4(-2 ± i)) / 2

Now, divide both terms in the numerator by 2 to simplify further:

x = 2(-2 ± i)

So, the complex solutions for the equation x^2 + 8x + 20 = 0 are:

x = 2(-2 + i) and x = 2(-2 - i)

In standard form:

x = -4 + 2i and x = -4 - 2i

Answer 2

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

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=(-b+√(b^2-4ac))/(2a)\\and\\x_2=(-b-√(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=(-8+√(8^2-4.1.20))/(2.1)=(-8+√(64-80))/(2)\\\\x_1=(-8+√(-16))/(2)=(-8+√(-1)√(16))/(2)\\\\x_1=(-8+i4)/(2)\\\\x_1=-4+2i`

Remember that:
`√(-1)=i`

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

`x_2=(-8-√(8^2-4.1.20))/(2.1)=(-8-√(64-80))/(2)\\\\x_2=(-8-√(-16))/(2)=(-8-√(-1)√(16))/(2)\\\\x_2=(-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.