Help please, good math students!!!!!!!!!!!!!!!!!!!!!!?

Question

Solve by completing the square: y = 2x^ - 4x + 24
^ stands for squared. Also as ahint I know the answer involves imaginary numbers so if you dont know those dont bother. Sorry but I am really stressed out right now, math exam tommorow! So if you could please explain it step by step this is the only part I dont get. Thanks!

Answer 1

y = 2x^ - 4x + 24
now for completing the square, make the y a zero and put it on the other side (I like doing it that way)
2x^ - 4x + 24 = 0
Now, I always like to move that integer with no variable to the other side, so subtract it from both sides
2x^ - 4x = -24
When completing the square, you cannot have a coefficient on the x that is being squared - in this case it is "2", so divide everything by 2 to get rid of it
x^ - 2x = -12
Now, you are going to have to divide the middle coefficient by two (it is -1, so divide by two and you get -.5) and then square that number. This will give you what you have to add to both sides
x^ - 2x + (-1)^ = -12 + (-1)^
-1^ is 1, so you have
x^ - 2x + 1 = - 12 + 1
now, when you have something like this:
Ax^ - Bx + C and it is a perfect square, it comes out to
(x - b/2)(x - b/2)
so, you would have
(x - 1) (x - 1) = - 11
that is the same as
(x-1)^ = - 11
now, square root both sides and you get
x-1= √-11
now, add one
x = √-11 + 1
but, since when you square negative numbers, they come out positive, this is ALSO an answer:
x = -√-11 + 1

However, you can't square root a negative, because no number times itself will give you a negative!
So, you can pretend the negative wasn't there by timesing the equation by i (imaginary number of negative square roots)

x= 1 ± i√11

± means plus or minus, so it is either
x= 1 - i√11
x= 1 + i√11

Answer 2

Set y = 0 , and then set the equation so that all the x's are on one side

2x^ -4x = -24.

Divide each side by the factor of x^ to get

x^ - 2x = -12.

Now complete the square of the left hand side of the
equation.

For the general case (x + a)^ = x^ + 2ax + a^ , so that
substituting -2 for 2a, we know that a = -1, and a^ = 1

So add 1 to each side of the equation:

x^ -2x +1 = -12 + 1

x^ -2x +1 = -11

(x-1)^ = -11

x-1 = + / - i * sqrt (11)
x = 1 +/- i * sqrt(11)

I think that will get you where you need to be.

Bob

Answer 3

Finding the roots of this equation:
2x²-4x+24=0
x²-2x+12=0
x²-2x+1=-11
(x-1)²=-11
x-1=±i√11
x=1±i√11

Answer 4

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Answer 5

Here is the solution:
y = 2x* - 4x + 24

here; the equation can be simplified as
y = x* - 2 x + 12

here a = 1, b= -2 & c = 12

using the formula

y = [- b +- (sq.rt b* - 4 a c)]/2a

we get
y = (1 + 2i sq.rt. 11) & (1 - 2i sq.rt. 11)

Hope this clarifies your doubt.

Answer 6

if your looking for the zeros of the problem use this formula.

-b + or - the square root of b^ - 4(a)(c) over 2a

Answer 7

y=2x^ - 4x + 24
plug into quadratic formula
-b +/- sqrt( b^ - 4ac )
--------------------------- = x
2 a

once u do, u get:
1+ 3.3166i
1- 3.3166i
i=imaginary number
mooooo!

Answer 8

I WOULD SAY 24

Answer 9

y= 2x*-4x+24
ummm.. dont know

Answer 10

sorry, i haven't learned that yet, wish i could help