# Quadratic Equations in Vertex Form

1. Jul 21, 2009

### alex2256

1. The problem statement, all variables and given/known data
Find the roots and the vertex for 3x^2 + 5x - 2

2. Relevant equations
ax^2 + bx + c
a(x - h) + k

3. The attempt at a solution
OK, this is what I attempted.

3x^2 + 5x - 2 = 0
3x^2 + 6x - x - 2 = 0
3x(x+2) - 1(x+2) = 0
x = 1/3 and x = -2

OK, so I found the roots, however, I am having difficulty understanding finding the vertex of THIS problem. The book says the vertex should be (-5/6, -4 1/12), so here's what I did.

3x^2 + 5x - 2 = 0
x^2 + 5/3x - 2/3 = 0
x^2 + 5/3x - (25/36) - (2/3 - 25/35) = 0
(x + 5/6)^2 - 49/36

therefore the vertex is -5/6 and -49/36; obviously I didn't get -4 1/12.

Can somebody help me and explain to me what's going on?

2. Jul 21, 2009

### Office_Shredder

Staff Emeritus
Yup, it's pretty simple. You went from

x2+5/3x-25/36 (emphasis mine EDIT: Actually, a bold minus sign looks no different than a regular one...)

to

(x+5/6)2

if yo factor this out, you'll see you get a +25/36. So your constant term is off because you put the negative 25/36 into the complete the square portion instead of the positive part