Simplify your answer as much as possible.
(If there is more than one solution, separate them with commas.)
solve (t+6)^2-45=0 where is a real number?
2008-07-12 7:58 pm
回答 (4)
2008-07-12 8:03 pm
✔ 最佳答案
(t + 6) = ± √ 45 (t + 6) = ± 3√5
t = - 6 ± 3√5
t = - 6 + 3√5 , t = - 6 - 3√5
2008-07-12 8:08 pm
(t + 6)^2 - 45 = 0
(t + 6)^2 = 45
t + 6 = ±√45
t + 6 = ±6.7 (approx.)
t + 6 = 6.7
t = 6.7 - 6
t = 0.6
t + 6 = -6.7
t = -6.7 - 6
t = -12.7
∴ t = -12.7 , 0.6
(t + 6)^2 = 45
t + 6 = ±√45
t + 6 = ±6.7 (approx.)
t + 6 = 6.7
t = 6.7 - 6
t = 0.6
t + 6 = -6.7
t = -6.7 - 6
t = -12.7
∴ t = -12.7 , 0.6
參考: Square Root Calculator
http://www.math.com/students/calculators/source/square-root.htm
2008-07-12 8:06 pm
[1] (t+6)^2 becomes equal to 45.
[2] that is, (t+6)^2 = 45
[3] taking the square root of both sides gives
[4] t+6 = +- sqrt(45)
[5] sqrt(45) simplifies to 3sqrt(5)
[6] finally, t = -6 +- 3sqrt(5)
[2] that is, (t+6)^2 = 45
[3] taking the square root of both sides gives
[4] t+6 = +- sqrt(45)
[5] sqrt(45) simplifies to 3sqrt(5)
[6] finally, t = -6 +- 3sqrt(5)
2008-07-12 8:00 pm
i think 0
收錄日期: 2021-05-01 10:48:56
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