Schoolwork Help Thread

[braix]

ok thanks ^^

[InsigTurtle]

Quick question.
How do you find complex roots, like in x^2-x+1=0?
I tried using the quadratic formula and I got x=(1 +- i sqrt -3)/2, but Wolfram Alpha says it's the cube root of -1 and -(-1)^2/3...
I'm confused?

And the cube root of -1 is 1, so isn't that a real number? Even more confused...

[Latios212]

You and Wolfram Alpha are both correct Although I'm not quite sure why WA writes it that way.

-1 actually has three cube roots, -1 and the two complex numbers you listed. Try cubing them!

[InsigTurtle]

Life was so much simpler when I only had to worry about one dimension of numbers
Thanks!

[mikey]

it's been like 2 years since I've done this stuff, I gotta get back into it!
although I kinda prefer stats

[Yugi]

For the parabola y = (x-2)to the power of two - 3, would the vertex be (2, 3) or (-2,-3)

[TheMarioPianist]

For the parabola y = (x-2)to the power of two - 3, would the vertex be (2, 3) or (-2,-3)

Actually, you are halfway right on both. The vertex is actually (2, -3). It's easier to visualize this if you add that 3 over to the other side. Your equation would then be:
y+3=(x-2)^2

To get the points, you should think this.
For the x coordinate: What do I subtract 2 from to get 0?
For the y coordinate: What do I add 3 to to get 0?

Hope that helped!

[Clanker37]

Or alternatively fun with calculus!

Assuming the equation is y = ((x-2)^2) - 3
Then dy/dx = 2(x-2)
Which expands to dy/dx = 2x - 4
Let dy/dx = 0, since at the vertex the gradient will be 0 as it is a turning point.
Hence, 0 = 2x - 4
Rearranges to 4 = 2x
Therefore x = 2

[blueflower999]

Alternatively the vertex of a parabola can be found by -b/2a

(X-2)^2 - 3 foils out into

X^2 - 4X -1, so b=-4 and a=1

4/2 = 2, so that's the X coordinate. Plug that back into the equation and you'll see that the vertex is (2, -3).

[mikey]

all of y'alls complex math then blue brings it back with some lame algebra

[Yugi]

I also have no clue what the x-intercept for this is, apparently its the square root of -7 but that makes no sense at all

[Latios212]

The x-intercept is where the graph crosses the x-axis. To find the x-intercept, set y=0 and solve for the x-coordinate(s).

[Bubbles]

What Latios said

Here's a workthrough but try it yourself first (no peeking!!)

Spoiler

y=(x-2)^2 -3
(0) = (x-2)^2 -3   ---  add three
3 = (x-2)^2  ---  take the square root of both sides (don't forget the square root of 3 can be positive or negative)
±√3 = x-2  ---  add 2
x = 2 ± √3
so

x = ~3.732
or
x = ~0.267

Idk where you got √-7 from, since thats not even a real number

[close]

[braix]

(no peeking!!)[/spoiler]

whoops

[mikey]

X EQUALS NEGATIVE B
PLUS OR MINUS THE SQUARE ROOT
OF B SQUARED MINUS FOUR A C
ALL OVER TWO A