Asked by drake
((x)(x))/(3-x) = 1.8x10^-5
how do I solve for x
how do I solve for x
Answers
Answered by
Damon
multiply bout sides by (3-x)
x^2 = 1.8x10^-5 (3-x)
x^2 + 1.8 * 10^-5 x - 5.4 *10^-5 = 0
plain old quadratic now
x^2 = 1.8x10^-5 (3-x)
x^2 + 1.8 * 10^-5 x - 5.4 *10^-5 = 0
plain old quadratic now
Answered by
oobleck
x^2/(3-x) = 1.8*10^-5
x^2 = 5.4*10^-5 - 1.8*10^-5 x
x^2 + 1.8*10^-5 x - 5.4*10^-5 = 0
I guess the quadratic formula is the best way to handle that. I get
x = -0.00735747, 0.00733947
or, more exactly,
x = (-1.8*10^-5 ±√((1.8*10^-5)^2+21.6*10^-5))/2
= (-0.000018±√(3.24*10^-10 + 21.6*10^-5))/2
= (-0.00018±√.000216000324)/2
= (-0.00018±0.0146969)/2
as above
x^2 = 5.4*10^-5 - 1.8*10^-5 x
x^2 + 1.8*10^-5 x - 5.4*10^-5 = 0
I guess the quadratic formula is the best way to handle that. I get
x = -0.00735747, 0.00733947
or, more exactly,
x = (-1.8*10^-5 ±√((1.8*10^-5)^2+21.6*10^-5))/2
= (-0.000018±√(3.24*10^-10 + 21.6*10^-5))/2
= (-0.00018±√.000216000324)/2
= (-0.00018±0.0146969)/2
as above
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