Am I right?

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Find the vertex of f(x) = 5x2 + 2x + 1. (-2±√(2^2-4*5*1))/(2*5) (-2±√(-16))/10 = (-2±-4)/10 = 2/10 and -6/10
no you are incorrect.... it would seem that you have found the x intercepts or the zeros for the quadratic
Oh, I mixed up the problem, that is what I was trying to find, LOL. Sorry, working all day on Algebra.

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Other answers:

I was trying to find the zeroes. Is it right?
if this is algebra the 1st step is to find the line of symmetry for the parabola, as the vertex is on the line of symmetry so for a parabola \[y=ax^2 + bx + c\] the line of symmetry is \[x = \frac{-b}{2 \times a}\] you have b = 2 and a = 5 when you get the value for x, substitute the x value into the equation to find the minimum value these to values will get the vertex... as a point
  • phi
2/10 and -6/10 you probably should simplify these to 1/5 and -3/5
Yes, I did. So it's right?
  • phi
oh, one subtlety! \[ \frac{-2 \pm \sqrt{-16}} {10 } \]
  • phi
the square root of -16 is sqrt(16) * sqrt(-1) or 4 i (we use i for the square root of -1)
So, I put i after the top negative number?
  • phi
one root is \[ \frac{-1}{5} + \frac{2}{5} i \] the other is \[ \frac{-1}{5} - \frac{2}{5} i \]
You mean 3/5?
  • phi
\[\frac{-2 \pm \sqrt{-16}} {10 } \\ \frac{-2 \pm 4i }{10 }\\ \frac{-2}{10} \pm \frac{4}{10} i \]
  • phi
and finally \[ - \frac{1}{5} \pm \frac{2}{5} i \]
  • phi
you cannot combine the term with "i" with the "real" term the term with "i" is called the imaginary part and the other part is called the "real part"
But, isn't the square root of -16, -4?
  • phi
-4*-4 = 16 (which is obviously not -16) people were very confused about this a few hundred years ago. they finally decided the only way to get -16 was to use \( \sqrt{-1} \) and then \[ \sqrt{-1} \cdot 4 \cdot \sqrt{-1} 4 = \sqrt{-1} \cdot \sqrt{-1} 16 = -1\cdot 16=-16\]
Ok, sorry, not very good at math, I'm a history and English guy.
  • phi
just remember \( \sqrt{-16} = 4i \)
Ok, thanks!!!

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