Question

Am I right?

Answer 1

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

Answer 2

no you are incorrect.... it would seem that you have found the x intercepts or the zeros for the quadratic

Answer 3

Oh, I mixed up the problem, that is what I was trying to find, LOL. Sorry, working all day on Algebra.

Answer 4

I was trying to find the zeroes. Is it right?

Answer 5

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

Answer 6

2/10 and -6/10

you probably should simplify these to 1/5 and -3/5

Answer 7

Yes, I did. So it's right?

Answer 8

oh, one subtlety!
\[ \frac{-2 \pm \sqrt{-16}} {10 } \]

Answer 9

the square root of -16 is sqrt(16) * sqrt(-1)
or
4 i (we use i for the square root of -1)

Answer 10

So, I put i after the top negative number?

Answer 11

one root is
\[ \frac{-1}{5} + \frac{2}{5} i \]
the other is
\[ \frac{-1}{5} - \frac{2}{5} i \]

Answer 12

You mean 3/5?

Answer 13

\[\frac{-2 \pm \sqrt{-16}} {10 } \\ \frac{-2 \pm 4i }{10 }\\
\frac{-2}{10} \pm \frac{4}{10} i \]

Answer 14

and finally
\[ - \frac{1}{5} \pm \frac{2}{5} i \]

Answer 15

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"

Answer 16

But, isn't the square root of -16, -4?

Answer 17

-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
\[ \sqrt{-1} \cdot 4 \cdot \sqrt{-1} 4 = \sqrt{-1} \cdot \sqrt{-1} 16 = -1\cdot 16=-16\]

Answer 18

Ok, sorry, not very good at math, I'm a history and English guy.

Answer 19

just remember \( \sqrt{-16} = 4i \)

Answer 20

Ok, thanks!!!