Started by Oct 15 2001 03:43 PM

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6 replies to this topic

Posted 15 October 2001 - 03:43 PM

I don't know where my head was back in math class, perhaps up something else, but now I'm regretting not paying attention.
Gotta find out when certain equations = 0
e.g.
x(x + 3) = 0 when x = 0 or x = -3
Easy enough, but what about 6(x^2 - x - 6)
(I figured out it was x = 3 & -2)
So what was the method for finding these things?
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- outRider -
Edited by - outRider on October 15, 2001 10:51:24 PM

Posted 15 October 2001 - 03:52 PM

There are a couple ways of doing this. First of all, you can factor the equation. 6(x^2 - x - 6) is equal to 6(x+2)(x-3), and any part of the equation equaling 0 makes the whole thing 0, so x can equal either -2 or 3. Also, there is the quadratic equation, which is (-b +or- sqrt(b^2-4ac))/2a for any equation ax^2 + bx + c = 0. Hope this helps!

Posted 15 October 2001 - 04:01 PM

Remember the quadratic equation?

You previous post said you can't figure out 6(x^2 - x - 6) = 0

Ok this is how you do it

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

So, you get 6((1 +/- sqrt(1 + 24)) / (2a))

Simplify you get

6((1 +- 5) / 2)

so the answers you get are

-2 and 3

The Quadratic Equation at its finest :-)

You could have also factored everyting into 6(x+2)(x-3) to get the same answer; the quadratic is better for more complex things.

Edited by - Viscous-Flow on October 15, 2001 11:02:13 PM

You previous post said you can't figure out 6(x^2 - x - 6) = 0

Ok this is how you do it

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

So, you get 6((1 +/- sqrt(1 + 24)) / (2a))

Simplify you get

6((1 +- 5) / 2)

so the answers you get are

-2 and 3

The Quadratic Equation at its finest :-)

You could have also factored everyting into 6(x+2)(x-3) to get the same answer; the quadratic is better for more complex things.

Edited by - Viscous-Flow on October 15, 2001 11:02:13 PM

Posted 15 October 2001 - 04:10 PM

ahhh, i LOVE the quadratic equation

it made me no longer have to factor those damn things

--- krez (krezisback@aol.com)

it made me no longer have to factor those damn things

--- krez (krezisback@aol.com)

Posted 15 October 2001 - 04:26 PM

I remember the quadratic equation but it doesn''t always give me meaningful answers. The reason I''m doing this is because I have a function g(x) = 3x^4 - 16x^3 + 6x^2 + 72x + 8

g`(x) = 12x^3 - 48x^2 + 12x + 72

To find the critical numbers I''d have to find all the values for x that the equation = 0...

using the quadratic equation after getting it in the form ax^2 + bx + c gets me into decimals and shit... bah, I''ll just factor.

Calculus blows when you have to take it in a CS program.

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- outRider -

g`(x) = 12x^3 - 48x^2 + 12x + 72

To find the critical numbers I''d have to find all the values for x that the equation = 0...

using the quadratic equation after getting it in the form ax^2 + bx + c gets me into decimals and shit... bah, I''ll just factor.

Calculus blows when you have to take it in a CS program.

------------

- outRider -

Posted 15 October 2001 - 04:27 PM

Basically the same as using the quadratic formula is completing the square. You have an equation of the form a*x^2+b*x+c=0. You can divide through by a to get x^2+b*x/a+c/a. Then to complete the square you add and subtract one half the x term coefficent squared (x^2+b*x/a+b^2/4a^2)+c/a-b^2/4a^2=0 which is the same as (x+b/2a)^2+c/a-b^2/4a^2 which is the same as (x+b/2a)^2+(4ac-b^2)/4a^2=0. Now to solve for x you move the constant to the right hand side, (x+b/2a)^2=(b^2-4ac)/4a^2, take the square root, x+b/2a=+/-sqrt(b^2-4ac)/2a, and finally subtract b/2a to get the quadratic formula x=-b/2a+/-sqrt(b^2-4ac)/2a or (-b+/-sqrt(b^2-4ac))/2a.

I always have trouble remembering the quadratic formula and find it much easier to remember how to derive it.

I always have trouble remembering the quadratic formula and find it much easier to remember how to derive it.

Posted 15 October 2001 - 05:14 PM

As has been previously mentioned, homework (or school, as opposed to Game Development related) questions should not be posted to this site. The answer to the above problem can be found in ANY elementary calculus text book. It only takes a walk to the library to look it up if you don''t own one yourself. Discovering the answer for yourself gains you far more understanding than someone telling it to you.

Regards,

Timkin

Regards,

Timkin