2007-12-02 13:35:08 · 9 answers · asked by Corey 1 in Science & Mathematics ➔ Mathematics
2 y ² = - 8
y ² = - 4
y ² = 4 i ²
y = ± 2 i
2007-12-03 04:07:06 · answer #1 · answered by Como 7 · 2⤊ 1⤋
You want to get "y" by itself on one side of the equation.
4^2 + 8 = 2y^2
2y^2 + 8 = 0 (I subtracted the 2y^2 from both sides)
2y^2 = -8 (I subtracted 8 from both sides)
y^2 = -4 (I divided both sides by 2)
y = sqrt(-4) (I took the square root of both sides)
y = 2 sqrt (-1) (I took the square root of four and left the -1 in the radical)
y = 2i (i is the exact same thing as the square root of -1)
Hope that helps!
2007-12-02 21:40:03 · answer #2 · answered by Sage B 4 · 1⤊ 0⤋
4y^2+8=2y^2
Subtract 2y^2 from both sides:
2y^2+8=0
2(y^2+4)=0
y^2=-4
There is no solution because you cannot square out a negative number.
2007-12-02 21:39:32 · answer #3 · answered by Anonymous · 0⤊ 1⤋
4y^2 + 8 = 2y^2
4y^2 - 2y^2 + 8 = 0
2y^2 + 8 = 0
2y^2 = -8
(2y^2) / 2 = -8 / 2
y^2 = -4
square root (y^2) and -4
y= (-4)^(1/2)
either you can look at it and say no solution, or you can say
y = 4i
2007-12-02 22:27:15 · answer #4 · answered by Il fuoco di furia 2 · 1⤊ 1⤋
4y^2 - 2y^2 = -8
2y^2= -8 ( divide both sides by 2)
y^2= -4 ( take sqrt. of both sides)
y= sqrt.(-4) (simplify)
y= 2i
i= sqrt. of negative 1
2007-12-02 22:01:34 · answer #5 · answered by W 2 · 1⤊ 1⤋
4y2 = 2y + 8
4y2 - 2y = 8
2y2 = 8
y2 = 4
y = 2
2007-12-02 21:38:02 · answer #6 · answered by Sarah 4 · 0⤊ 1⤋
use FOILing
4y2 -2y + 8 = 0
(2y + 2)(2y - 4) = 0
solve for y
and there you have it.
2007-12-02 21:41:18 · answer #7 · answered by Anna L 1 · 1⤊ 1⤋
the answer is: 8y to the fourth power =8
2007-12-02 21:39:17 · answer #8 · answered by Anonymous · 0⤊ 1⤋
there is no answer.
'unsolvable'
2007-12-02 21:40:05 · answer #9 · answered by Anonymous · 1⤊ 0⤋