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la_nena26
07-11-2006, 05:00 PM
25x^2 - 64 = 0

tkhunny
07-11-2006, 05:02 PM
You simply MUST learn to recognize a "Difference of Squares".

25x^2 - 64 = (5x+8)(5x-8)

If all else fails, you can ALWAYS use the quadratic formula.

25x^2 + 0x - 64 ==> a = 25, b = 0, c = 64

la_nena26
07-11-2006, 05:17 PM
You simply MUST learn to recognize a "Difference of Squares".

25x^2 - 64 = (5x+8)(5x-8)

If all else fails, you can ALWAYS use the quadratic formula.

25x^2 + 0x - 64 ==> a = 25, b = 0, c = 64

is it the same if you are adding?

tkhunny
07-11-2006, 05:31 PM
is it the same if you are adding?No. That doesn't work. Both sum and difference of CUBES can be factored.

la_nena26
07-11-2006, 06:11 PM
so if i got 16n^2+18n=0
i know i have to find the square of 16 which is 4
(4n+ )(4n+ ) what do i do next with the 18?

stapel
07-11-2006, 06:46 PM
so if i got 16n^2+18n=0
In future, please post new questions as new threads, not as replies to old questions where they may be overlooked (or answered incorrectly). Thank you.

To solve a quadratic by factoring, first factor whatever is common to all of the terms. In the case of 16n<sup>2</sup> + 18n, there is a 2n common to both terms.

Eliz.

jonboy
07-11-2006, 07:32 PM
We have: \L 16n^2+18n=0

As stapel said we have a common term of \L 2n

So: \L 2n(8n+9)=0

la_nena26
07-11-2006, 08:37 PM
Thanks everybody! :D

Denis
07-12-2006, 12:05 AM
25x^2 - 64 = 0

Worth a shot at isolating x^2:
25x^2 = 64
x^2 = 64 / 25
x = +-sqrt(64 / 25)
x = +- 8/5