16y^2 = 25
2007-07-08 09:31:45 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1. subtract 25 from both sides to make your equation equal to zero & simplify
16y^2 (-25) = 25 (-25)
16y^2 - 25 = 0
2. determine the factors of 16 & 25. **now if you know that 16 & 25 are perfect squares then this problem is really easy if not then you will have to do trial & error to see which numbers work.
(4y-5)(4y+5) --> --> --> --> YOUR ANSWER
**don't forget your positives & negatives rules when multiplying**
2007-07-08 11:56:02 · answer #1 · answered by bnhawk03 3 · 0⤊ 0⤋
16y^2 - 25 = 0
Now it is a difference of squares.
(4y - 5) (4y + 5) = 0
If you are solving,
y = 5/4, or -5/4
2007-07-08 09:34:58 · answer #2 · answered by de4th 4 · 0⤊ 0⤋
(16/25)y^2 = 1
y^2 = 25/16
y^2 = (5/4)^2
y^2 - (5/4)^2 = 0
(y - 5/4)(y + 5/4) = 0
2007-07-08 09:37:37 · answer #3 · answered by Amit Y 5 · 0⤊ 0⤋
16y^2=25
or 16y^2-25=0
or (4y)^2-(5)^2=0
or (4y-5)(4y+5)=0 ans
2007-07-08 09:39:10 · answer #4 · answered by MAHAANIM07 4 · 0⤊ 0⤋
take the square root of both sides and you get
4y=5
y=5/4
2007-07-08 09:35:22 · answer #5 · answered by cerah_micah 3 · 0⤊ 0⤋