help me solve for y: x^2 - 30x + 2y^2 + 125 = 0?
2009-05-12 7:47 am
can anyone help me solve for y.. I set all the xs to 0.. but you cant take the sqrt of a negative number.. confused.. thanks alot.
回答 (5)
2009-05-12 10:28 pm
✔ 最佳答案
(x- 15)^2 + 2 y^2 =100= 10^2this is an equation of a circle.
It is an elliptic equation.Divided it by 2 on both sides.
2009-05-12 10:04 am
x^2 - 30x + 2y^2 + 125 = 0
2y^2 = -x^2 + 30x - 125
y^2 = -(x^2 - 30x + 125)/2
y^2 = -(x^2 - 5x - 25x + 125)/2
y^2 = -[(x^2 - 5x) - (25x - 125)]/2
y^2 = -[x(x - 5) - 25(x - 5)]/2
y^2 = -[(x - 5)(x - 25)]/2
y = 屉{-[(x - 5)(x - 25)]/2}
y = ±iâ{[(x - 5)(x - 25)]/2}
2y^2 = -x^2 + 30x - 125
y^2 = -(x^2 - 30x + 125)/2
y^2 = -(x^2 - 5x - 25x + 125)/2
y^2 = -[(x^2 - 5x) - (25x - 125)]/2
y^2 = -[x(x - 5) - 25(x - 5)]/2
y^2 = -[(x - 5)(x - 25)]/2
y = 屉{-[(x - 5)(x - 25)]/2}
y = ±iâ{[(x - 5)(x - 25)]/2}
2009-05-12 8:15 am
2y^2 = -(x^2+30x+125) = -(x^2+30x+225 – 100) = -(x+15)^2 + 100.
2y^2 = 100 if x+15 = 0 (as y^2 > 0, x+15 =< 10). => y^2 (max) = 50
=> y(max) = 5â2
2y^2 = 100 if x+15 = 0 (as y^2 > 0, x+15 =< 10). => y^2 (max) = 50
=> y(max) = 5â2
2009-05-12 8:05 am
Are you studying circles. If you are when x = 15 y= +or - 5 sqrt 2 or when y is 0 x is 25
2009-05-12 8:01 am
2y^2=-x^2+30x-125
y = + & - \/''''(-x^2''+30x-125)/2'''''''''''''''
Sorry I CAN NOT show radicals by my keybord.
y = + & - \/''''(-x^2''+30x-125)/2'''''''''''''''
Sorry I CAN NOT show radicals by my keybord.
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