2006-08-11 05:04:38 · 7 answers · asked by corey c 1 in Science & Mathematics ➔ Mathematics
y^2 + 4y = 4
y^2 + 4y + 4 = 8
(y + 2)^2 = 8
y + 2 = +/- 2 sqrt(2)
y = -2 +/- 2 sqrt(2)
2006-08-11 05:31:16 · answer #1 · answered by bpiguy 7 · 0⤊ 0⤋
Assuming y2 is y^2 (y-squared), the equation isn't: merely improve the words contained in the brackets and also you get: y^2 + 4y - y^2 = 6 Which simplifies to: 4y = 6 which isn't a quadratic equation. yet when y2 isn't y-squared yet truly 2y (i.e. 2 more effective with the help of y) then the equation is quadratic: y^2 + 2y = 6. desire this facilitates. suited answer? :P
2016-11-24 20:12:20 · answer #2 · answered by speelman 4 · 0⤊ 0⤋
4 - 4y - y^2 = 0
-y^2 - 4y + 4 = 0
-(y^2 + 4y - 4) = 0
y^2 - 4y - 4 = 0
y = (-b ± sqrt(b^2 - 4ac))/(2a)
y = (-(-4) ± sqrt((-4)^2 - 4(1)(-4)))/(2(1))
y = (4 ± sqrt(16 + 16))/2
y = (4 ± sqrt(32))/2
y = (4 ± sqrt(16 * 2))/2
y = (4 ± 4sqrt(2))/2
ANS : 2 + 2sqrt(2) or 2 - 2sqrt(2)
2006-08-11 13:01:31 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
-(y^2+4y-4)=0 values are -4+4(2)^1/2/2 and -4-4(2)^1/2//
=-2+2rt2 and -2-2rt2
2006-08-11 05:15:23 · answer #4 · answered by raj 7 · 0⤊ 0⤋
factorize it by using the quadratic formula:
(-b +- sqrt(D))/2a, D is the determinant formula I gave you on the other question.
or using the completing the squares method
2006-08-11 05:22:52 · answer #5 · answered by Lie Ryan 6 · 0⤊ 1⤋
corey, what part of your math assignment are you doing? (Just posting problems is not doing your assignment). I thought you knew that...
2006-08-11 17:38:37 · answer #6 · answered by ronw 4 · 0⤊ 1⤋
0.828 or-4.828 (correct upto 3 digit)
2006-08-11 05:14:34 · answer #7 · answered by Som 1 · 0⤊ 0⤋