Question

What are the zeros of the function f (x)=x^2+5x+5 written in simplest radical form​

Answer 1

The zeros are

`x1=\frac{-5+√(5)} {2}` and
`x2=\frac{-5-√(5)} {2}`

Explanation:

we know that

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`f(x)=x^(2) +5x+5`

To find the zeros equate the function to 0

`x^(2) +5x+5=0`

so

`a=1\\b=5\\c=5`

substitute in the formula

`x=\frac{-5(+/-)\sqrt{5^(2)-4(1)(5)}} {2(1)}`

`x=\frac{-5(+/-)√(5)} {2}`

`x1=\frac{-5+√(5)} {2}`

`x2=\frac{-5-√(5)} {2}`