the answer
factor and solve
5y(squared) + 2y(squared) - 3 = 0
2007-02-23 14:58:52 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I'm going to assume you mean
5y^2 + 2y - 3 = 0
(I agree with the answer that the second term isn't supposed to be squared).
Factor; through trial and error you should get
(5y - 3)(y + 1) = 0
meaning y = 3/5 or -1
2007-02-23 15:04:57 · answer #1 · answered by Puggy 7 · 0⤊ 0⤋
5y^2 + 2y^2 - 3 = 0 => 7y^2 = 3 => y^2 = 3/7 => y = +/- sqrt(3/7). Are you sure this was the problem?? And not 5y^2 + 2y - 3 = 0??
2007-02-23 23:07:57 · answer #2 · answered by flyfisher_20750 3 · 0⤊ 0⤋
5y^2 + 2y^2 - 3 = 0
7y^2 - 3 = 0
7y^2 = 3
y^2 = 3/7
y = square root of 3/7
2007-02-23 23:03:59 · answer #3 · answered by the DoEr 3 · 0⤊ 0⤋
if ...... 5y(squared) + 2y(squared) - 3 = 0
7y^2 - 3 = 0
y = sqrt 3/7
if .........5y(squared) + 2y - 3 = 0
5y^2 + 2y - 3 = 0
(5y -3)(y+1) = 0
y = 3/5 or -1
2007-02-23 23:07:18 · answer #4 · answered by Anonymous · 0⤊ 0⤋
5y(squared) + 2y(squared) - 3 = 0
7y(squared)=3
y(squared)=7/3
y=plus or minus sqrt of (7/3)
y=plus or minus sqrt of (21)/3
2007-02-23 23:08:18 · answer #5 · answered by dave 2 · 0⤊ 0⤋
5y^2+2y^2-3=0
7y^2-3=0
7y^2=3
y^2=3/7
y=+/-â(3/7)=+/-0.65465367070797714379829245624686
unless you meant 5y^22+y-3=0
(5y-3)(y+1)=0
5y-3=0
5y=3
y=0.6
y+1=0
y=-1
y=0.6, -1
2007-02-24 00:01:23 · answer #6 · answered by yupchagee 7 · 0⤊ 0⤋
I bet the second squared isnt supposed to be there.
2007-02-23 23:02:04 · answer #7 · answered by Anonymous · 0⤊ 0⤋