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

1 Answer
George C.
Feb 1, 2018

#x = -5/2+-sqrt(5)/2#

Explanation:

Given:

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

Method 1 - Completing the square

Solve:

#0 = 4f(x)#

#color(white)(0) = 4(x^2+5x+5)#

#color(white)(0) = 4x^2+20x+20#

#color(white)(0) = (2x)^2+2(2x)(5)+25-5#

#color(white)(0) = (2x+5)^2-(sqrt(5))^2#

#color(white)(0) = ((2x+5)-sqrt(5))((2x+5)+sqrt(5))#

#color(white)(0) = (2x+5-sqrt(5))(2x+5+sqrt(5))#

So:

#2x = -5+-sqrt(5)#

Dividing both sides by #2#, we find:

#x = -5/2+-sqrt(5)/2#

Method 2 - Quadratic formula

Note that #f(x)# is in standard quadratic form:

#f(x) = ax^2+bx+c#

with #a=1#, #b=5# and #c=5#.

This has zeros given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-(color(blue)(5))+-sqrt((color(blue)(5))^2-4(color(blue)(1))(color(blue)(5))))/(2(color(blue)(1)))#

#color(white)(x) = (-5+-sqrt(25-20))/2#

#color(white)(x) = (-5+-sqrt(5))/2#

#color(white)(x) = -5/2+-sqrt(5)/2#

Related questions
Impact of this question
6760 views around the world
You can reuse this answer
Creative Commons License