math question

Completing the square looks good at a glance.

math man wrote:

how do you solve the equation Y=X(2-X), for X?

Well X can only equal 0 or 1 in a problem like that i believe...not sure how to do it off the top of my head.

math man wrote:

how do you solve the equation Y=X(2-X), for X?

factor it out to y=x^2-2x

i know there is an easier way, but only thing i can think of is qua equation. a=1 b=-2 c=0

2/2(1) +/- (sqrt(4-4(1)(0))/2(1)

= 1 +/- (sqrt(4))/2

= 1 +/- 2/2 = 1 +/- 1 = 0 and -2

x = 0,-2

whoops.. sorry... completely disregard what i just said.... I was thinking that it was set equal to 0.

I'm a retard

Mrr82 wrote:

math man wrote:

how do you solve the equation Y=X(2-X), for X?

Well X can only equal 0 or 1 in a problem like that i believe...not sure how to do it off the top of my head.

Yikes!

Where did you go to school?

X = 1 +/- sqrt(1-Y)

sally G. wrote:

math man wrote:

how do you solve the equation Y=X(2-X), for X?

factor it out to y=x^2-2x

i know there is an easier way, but only thing i can think of is qua equation. a=1 b=-2 c=0

2/2(1) +/- (sqrt(4-4(1)(0))/2(1)

= 1 +/- (sqrt(4))/2

= 1 +/- 2/2 = 1 +/- 1 = 0 and -2

x = 0,-2

By the way, it would be MINUS 2/2(1) +/- (sqrt(4-4(1)(0))/2(1), in case this is something the OP was looking for. That would yield 0, 2 which could have been guessed by looking at the original expression.

y=x(2-x)

y=2x-x^2

-x^2+2x-y=0

a=-1, b=2, c=-y

x=(-b(+or-)sqrt(b^2-4ac))/2a

x=(1(+or-)sqrt(4-4*-1*y))/-2

x=-0.5-0.5*(4+4y)^0.5

OR

x=-0.5+0.5*(4+4y)^0.5

Whoops, that wasn't the problem, I forgot b was negative already. Here we go:

= 1 +/- 2/2 = 1 +/- 1 = 0 and 2

kartelite wrote:

sally G. wrote:

factor it out to y=x^2-2x

i know there is an easier way, but only thing i can think of is qua equation. a=1 b=-2 c=0

2/2(1) +/- (sqrt(4-4(1)(0))/2(1)

= 1 +/- (sqrt(4))/2

= 1 +/- 2/2 = 1 +/- 1 = 0 and -2

x = 0,-2

By the way, it would be MINUS 2/2(1) +/- (sqrt(4-4(1)(0))/2(1), in case this is something the OP was looking for. That would yield 0, 2 which could have been guessed by looking at the original expression.

Don't listen to her. X has to be a function of Y. In the quadratic equation c should be equal to -Y. Plug 0 or -2 into the original equation and you'll get Y=0 or Y=8, not Y=Y as you should.

Really has anyone passed Algebra 1 yet?

You have got to be shitting me. Are you an idiot?

kartelite wrote:

X = 1 +/- sqrt(1-Y)

This is the answer, you were asked to solve for X not find the zeros.

wow, the answers are:

x = -(sqrt(1-y))- 1)

or

x = (sqrt(1-y))+ 1)

I guess if I need to show my work I can. I'm happy my college education is paying off

I. Newton wrote:

You have got to be shitting me. Are you an idiot?

What? I provided the answer didn't I? I'm a math grad student, while that doesn't preclude me from being an idiot, well...

math man wrote:

how do you solve the equation Y=X(2-X), for X?

In distributing the x, you would get 2x-x squared, meaning it's a quadratic equation.

There are three ways of solving quadratic equations: factoring, completing the square, and the quadratic formula. Regardless of which method you choose, the equation must be y=. Make y=0 and solve. Since it already was in factored form, this is the easiest way. Make both parts be equal to zero, so we have x=0 and 2-x=0. Solve the little equations and you get x=o and x=2.

These represent the x intercepts (where the graph crosses the x-axis). You can see this if you type the equation into y= in a graphing calculator. These are also known as the zeros or roots.

(Note: If you're taking algebra I, you likely will only use the factoring; completing the square and the quadratic formula usually are not covered until algebra II.)

Perhaps a reading comprehension class is in store?

it is quite shocking at times, isn't it kartelite?