Question 604794: I need to find the and approximate solutions to three decimal places for the equation x^2-5x+5=0 the solutions I have are (5,0,(-5,0) and a vertex at (0,5) but I know something incorrect
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! I need to find the and approximate solutions to three decimal places for the equation x^2-5x+5=0 the solutions I have are (5,0,(-5,0) and a vertex at (0,5) but I know something incorrect.
**
Since you have to find the vertex, write the equation in standard form for a parabola as follows:
y=x^2-5x+5
complete the square
y=(x^2-5x+25/4)+5-25/4
y=(x-5/2)^2-5/4
vertex: (5/2,-5/4)=(2.500,-1.250)
..
x-intercepts
set y=0
0=(x-5/2)^2-5/4
(x-5/2)^2=5/4
take sort of both sides
x-5/2=±√5/2
x=5/2±√5/2=2.500±1.118=3.618 and1.382
x-intercepts: 3.618 and1.382
|
|
|
