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

Question

asked Jul 18, 2020 86.3k views

1 Answer

YaTaras · Answer 1 · 2020-07-25T08:31:01+0000

Answer:

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

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

$x=\frac{-5(+/-)√(5)} {2}$

$x1=\frac{-5+√(5)} {2}$

$x2=\frac{-5-√(5)} {2}$