SOLUTION: ok so my question is how can you still use the "answer" you get from the Quadratic formula to find the vertex? And this is what i did... Problem: y= x^2-4x+6 i used the quadra

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: ok so my question is how can you still use the "answer" you get from the Quadratic formula to find the vertex? And this is what i did... Problem: y= x^2-4x+6 i used the quadra      Log On


   

Question 130120: ok so my question is how can you still use the "answer" you get from the Quadratic formula to find the vertex? And this is what i did...
Problem: y= x^2-4x+6
i used the quadratic formula...
x= 4(+or-) +sqrt%28+4%5E2-4%281%29%286%29+%29+ / 2(1)
and got x= 2 - 1.415i and x= 2 + 1.415i
how can i use those conjugate pairs to find the vertex?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
To find the x-coordinate of the vertex, simply average the two roots.


2+-+1.415i%2B+2+%2B+1.415i Add the two roots


%282%2B2%29+%2B%281.415i-+1.415i%29 Group like terms


4+-+0i Combine like terms


4 Simplify


So it turns out that when you add any two complex conjugates together you will get a real answer.


Now divide the answer by 2

4%2F2=2

So the x-coordinate of vertex is 2






note:
Remember the quadratic formula is

x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29 and the formula for the x-coordinate of the vertex is x=-b%2F%282a%29. Notice if you add the two roots the square root part cancels out to to give you -b%2F%282a%29. So this shows how the quadratic formula and the vertex are related.