Question

What are the zeros of the function f(x) = x2 + 5x + 5 written in simplest radical form?

Answer 1

Answer:

   x  =   $\frac{-5-\sqrt{5} }{2} , \frac{-5+\sqrt{5} }{2}$

Step-by-step explanation:

Given equation f(x) = x² + 5 x +5

The roots of the Quadratic equation   ax²+bx+c =0 are

           $x = \frac{-b-\sqrt{b^{2}-4 a c } }{ 2 a} , \frac{-b+\sqrt{b^{2}-4 a c } }{ 2 a}$

Given equation f(x) = x² + 5 x +5

comparing           ax²+bx+c =0 is

                a = 1 , b = 5 and c = 5

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

            x  =   $\frac{-5-\sqrt{5} }{2} , \frac{-5+\sqrt{5} }{2}$

             x  = -7.76 ,  x = -2.76 ,