3y^2 + 36 = 24
solve for y
2007-08-30 17:11:49 · 5 answers · asked by henry c 1 in Science & Mathematics ➔ Mathematics
3y^2 + 36 = 24
3y^2 + 12 = 0
y^2+4=0
y=2i,-2i
i=sqroot of -1 complex roots
2007-08-30 17:23:48 · answer #1 · answered by niki einstien 2 · 0⤊ 0⤋
Subtract 36 from each side.
You get 3y^2 = -12.
Divide by 3 and you get y^2= -4.
Take the square root and you get y = plus or minus 2i.
Thus y can be plus or minus 2i.
2007-08-31 00:24:01 · answer #2 · answered by Green Burger 1 · 0⤊ 0⤋
3y^2+36=24,Divide the whole equation by 3
y^2+12=8
y^2+12-8=8-8
y^2+4=0
y=2i,-2i
2007-08-31 00:31:31 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Divide by 3 to get: y^2 +12 = 8
subtract twelve : y^2 = -4
take square root : y = 2i ( an imaginary number)
Otherwise there is no REAL solution.
2007-08-31 00:18:57 · answer #4 · answered by james w 5 · 1⤊ 1⤋
3 y ² = - 12
y ² = - 4
y ² = 4 i ²
y = ± 2 i
2007-08-31 04:26:11 · answer #5 · answered by Como 7 · 1⤊ 0⤋