ok.. it says i need to put it into the ax^2+bx+c=o form ..
the problem is .. 49x^2=64
so i wrote ... 49x^2-64=o
now what do i do, bc only 7 goes into 49 evenly, and 8 into 64..but not both ??
2007-08-05 16:33:29 · 6 answers · asked by CHICKENBUTT 1 in Science & Mathematics ➔ Mathematics
If you have to put 49x^2-64=0 in ax^2+bx+c=0 form,it would be
49x^2+0x-64=0
The next step should be
49x^2+(56-56)x-64=0
or,49x^2+56x-56x-64=0
or,7x(7x+8)-8(7x+8)=0
or,(7x+8)(7x-8)=0
Therefore either 7x+8=0 or 7x-8=0
Hence x= -8/7 or 8/7
2007-08-05 16:45:14 · answer #1 · answered by Anonymous · 2⤊ 0⤋
49x² - 64 = 0
(7x - 8) (7x + 8) = 0
x = 8 / 7 , x = - 8 / 7
2007-08-06 03:05:50 · answer #2 · answered by Como 7 · 0⤊ 0⤋
You are right so far.
Now you have to use the difference of squares formula.
a^2 - b^2 = (a - b)(a + b)
In your case, a = 7x and b = 8
So you have:
(7x - 8)(7x + 8) = 0
x = ± 8/7
2007-08-05 23:37:12 · answer #3 · answered by whitesox09 7 · 0⤊ 0⤋
You are done. a = 49, b = 0, c = -64
2007-08-05 23:36:40 · answer #4 · answered by bozo 4 · 0⤊ 1⤋
a=49
b=0
c=64
Solution
49x^2=64
49x^2-64=0
(7x-8)(7x+8)=0
7x-8=0
x=8/7
or
7x+8=0
x=-8/7
2007-08-06 00:48:21 · answer #5 · answered by yenchin 2 · 0⤊ 0⤋
(7x + 8)(7x - 8)
x = -8/7 or 8/7
2007-08-05 23:36:45 · answer #6 · answered by Anonymous · 0⤊ 1⤋