I am lost: 25x^2 - 64 = 0

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
 
tkhunny said:
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
Click to expand...

is it the same if you are adding?
 
la_nena26 said:
is it the same if you are adding?
Click to expand...
No. That doesn't work. Both sum and difference of CUBES can be factored.
 
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?
 
la_nena26 said:
so if i got 16n^2+18n=0
Click to expand...
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.
 
We have: \(\displaystyle \L 16n^2+18n=0\)

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

So: \(\displaystyle \L 2n(8n+9)=0\)
 
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
 
You must log in or register to reply here.
Top