Question

What does the fundamental theorem of algebra state about the equation 2x^2−4x+16=0 ?

Answer 1

B

Explanation:

I took the test :)

Answer 2

option B

Explanation:

`2x^2-4x+16=0`

We need to solve this equation using quadratic formula

`x= (-b+-√(b^2-4ac))/(2a)`

a=2, b= -4, c=16

Plug in the values in the formula

`x= (-(-4)+-√((-4)^2-4(2)(16))/(2(2))`

`x= (4+-√(16-128))/(4)`

`x= (4+-√(-112))/(4)`

Simplify the square root. the value of square root (-1) = 'i'

`x= (4+-4i√(7))/(4)`

Now we divide by 4

`x= 1+-i√(7)`

So there are two complex roots. since the degree of polynomial is 2