anonymous
  • anonymous
Am I right?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
campbell_st
  • campbell_st
no you are incorrect.... it would seem that you have found the x intercepts or the zeros for the quadratic
anonymous
  • anonymous
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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anonymous
  • anonymous
I was trying to find the zeroes. Is it right?
campbell_st
  • campbell_st
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
  • phi
2/10 and -6/10 you probably should simplify these to 1/5 and -3/5
anonymous
  • anonymous
Yes, I did. So it's right?
phi
  • phi
oh, one subtlety! \[ \frac{-2 \pm \sqrt{-16}} {10 } \]
phi
  • phi
the square root of -16 is sqrt(16) * sqrt(-1) or 4 i (we use i for the square root of -1)
anonymous
  • anonymous
So, I put i after the top negative number?
phi
  • phi
one root is \[ \frac{-1}{5} + \frac{2}{5} i \] the other is \[ \frac{-1}{5} - \frac{2}{5} i \]
anonymous
  • anonymous
You mean 3/5?
phi
  • phi
\[\frac{-2 \pm \sqrt{-16}} {10 } \\ \frac{-2 \pm 4i }{10 }\\ \frac{-2}{10} \pm \frac{4}{10} i \]
phi
  • phi
and finally \[ - \frac{1}{5} \pm \frac{2}{5} i \]
phi
  • 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"
anonymous
  • anonymous
But, isn't the square root of -16, -4?
phi
  • 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 \[ \sqrt{-1} \cdot 4 \cdot \sqrt{-1} 4 = \sqrt{-1} \cdot \sqrt{-1} 16 = -1\cdot 16=-16\]
anonymous
  • anonymous
Ok, sorry, not very good at math, I'm a history and English guy.
phi
  • phi
just remember \( \sqrt{-16} = 4i \)
anonymous
  • anonymous
Ok, thanks!!!

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