View Full Version : Rather difficult maths equation

05-20-2004, 04:47 PM
Simplify: (2t)^2+(k-2^t)^2-(k-2t^2-1)^2

Please help me with this, I am practicing for a test, and I have been trying to solve this for hours!!!

05-20-2004, 05:32 PM
That's rather ugly. First expand everything out:
4t^2 + (k^2 - k2^t - k2^t + 2^2t) - (k^2 - 4kt^2 - 2k + 4t^4 + 4t^2 + 1)

Then combine like terms and you're done.
-(k2^(t+1)) + 2^2t + 4kt^2 + 2k - 4t^4 - 1

05-20-2004, 05:35 PM
I'm not gonna do it all for you, but expand out the powers:

4t^2 + (k-2^t)(k-2^t) - (k - 2t^2 - 1)(k - 2t^2 - 1)

and multply the contents within the braces. The rest is basic algebra and you should be able to group like terms together.

05-20-2004, 05:45 PM
I've been trying to work with the same methods as you. My facit says that the answer should be 2k-1.

I am starting to doubt that the facit is right. I've been trying your methods, but they haven't got me to the right answer. Either that, or there are some hidden conjugate rule or something like that that could make the whole process of solving much easier? Do you think that's possible?

05-20-2004, 06:15 PM
If you change the problem, then it's possible. Look at this:

It's very similar to your problem, and does simplify down to 2k-1. Are you sure you copied it right?

05-20-2004, 10:22 PM
we should have that math tag like they do in those calculus forums.

05-21-2004, 05:12 AM
Pianorain, you are indeed right.

The correct problem looks like this:

But I would like to know how you can simplify down to 2k-1, are there any shortcuts rather than expanding everything and then counting it together, which makes it very easy to lose a variable or two?

05-21-2004, 08:31 AM
One simplification you can make is to see that (k-2t^2) is a reccuring term - make it easier on yourself by substituting say (2t)^2 = a and (k-2t^2) = b which gives you the much
more friendly a + b^2 - (b - 1)^2 to simplify. From there on in it's pretty simple.