Question

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

Answer 1

Answer:

The zeros are

$x1=\frac{-5+\sqrt{5}} {2}$   and $x2=\frac{-5-\sqrt{5}} {2}$

Step-by-step 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

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

$x=\frac{-5(+/-)\sqrt{5}} {2}$

$x1=\frac{-5+\sqrt{5}} {2}$

$x2=\frac{-5-\sqrt{5}} {2}$