# Alg. 2 question

1. Mar 30, 2005

### woodworker101

If A=9^9999 + 9^-9999 and B= 9^9999 - 9^-9999.
Then find the value of A^2 -B^2 + 1.

would i just take the the numbers and square them to get the value. What should i do.

2. Mar 30, 2005

### joeboo

$$A^2 - B^2 = ( A+B )( A-B )$$

for any $A, B$
That should simplify it a great deal

3. Mar 30, 2005

### woodworker101

so would the values be 0 or 18. this question is confusing to me even when you did simplify it for me

4. Mar 30, 2005

### James R

Put p = 9^9999 and q= 9^-9999

Then A = p+q and B = p-q

You want:

$$A^2 + B^2 - 1 = (p+q)^2 + (p-q)^2 - 1 = (p^2 + q^2 + 2pq) + (p^2 +q^2 - 2pq) - 1$$
$$= 2p^2 + 2q^2 - 1 = 2(9^{9999}) + 2(9^{-9999}) - 1$$

5. Mar 31, 2005

### Curious3141

Unnecessarily complicated, and I'm afraid you read the question wrongly.

$$A = 9^{9999} + 9^{-9999}$$ and $$B = 9^{9999} - 9^{-9999}$$

$$A + B = (2)(9^{9999})$$ and $$A - B = (2)(9^{-9999})$$

$$A^2 - B^2 + 1 = (A + B)(A - B) + 1 = (4)(9^{9999})(9^{-9999}) + 1 = 4 + 1 = 5$$

6. Mar 31, 2005

### James R

I would argue that my solution is no more complicated than yours. You are, however, correct that I copied the question wrongly. My correct solution is:

$$A^2 - B^2 + 1 = (p+q)^2 - (p-q)^2 + 1 = (p^2 + q^2 + 2pq) - (p^2 +q^2 - 2pq) + 1$$
$$=4pq + 1 = 4(9^{9999})(9^{-9999}) + 1 = 4 + 1 = 5$$

So, we agree.

7. Mar 31, 2005

### woodworker101

they both seem right to me just that the letters are changed and makes it a little organized and not messy. So the right answer is 4+1 = 5.
Thanks for the help.

8. Mar 31, 2005

### woodworker101

I was wondering if you an explantation for what you did since my teachers requires how we got it instead of just stating a thoure (how ever you spell it) about how we got it. it is kind of stuiped but makes perfect sense to him.

9. Apr 1, 2005

### Curious3141

I feel it's needlessly complicated to expand out the square terms, you can factorise immediately without doing that. With your notation, it would simply be :

$$(p + q)^2 - (p - q)^2 + 1 = (p + q + p - q)(p + q - p + q) + 1 = (2p)(2q) + 1 = 4pq + 1$$

and I personally think that is simpler, IMHO. But let's not split hairs.

10. Apr 1, 2005

### James R

It's two ways to get to the same destination. I agree your method is perhaps neater, but requires more insight at the start, and so is perhaps not as direct as mine. We could argue over pedagogic value, too, but I'm happy to leave it here.

Gee, I have a big ego! :)