How do you solve t^2=6t+55?
回答 (10)
✔ 最佳答案
t ² - 6 t - 55 = 0
( t - 11 ) ( t + 5 ) = 0
t = 11 , t = - 5
t² = 6t + 55
t² – 6t – 55 = 0
t² + 5t – 11t – 55 = 0
(t + 5) – 11(t + 5) = 0
(t + 5)(t – 11) = 0
(t + 5) = 0 or (t – 11) = 0
t = – 5 or t = 11
move everything to one side:
t^2 - 6t - 55 = 0
factor if possible. If not use the quadratic formula.
(t-11)(t+5) = 0
t = 11 or t = -5
Since this is a quadratic equation we can put this is in standard form... something equal 0...
t^2 -6t -55=0
Now factor
(t-11)(t+5)=0
Use the zero property
t-11=0 and t+5=0
t=11 and t=-5
t^2-6t-55=0
(t+5)(t-11)=0
t= -5 t=11
t^2 - 6t - 55 =0
which factorises to
(t-11)(t+5)=0
so t=11 and t=-5
t^2-6t-55=0
t^2-11t+5t-55=0
t(t-11)+5(t-11)=0
(t-11)(t+5)=0
t=11 or t=-5
make it into a quadratic:
t^2 - 6t - 55 = 0
(t+5)(t-11) = 0
t= -5 or t= 11
Use the quadratic formula:
t^2 - 6t - 55 = 0
t = (6 +- sqrt(6^2 + 4*55))/2
= (6 +- sqrt(256))/2
= (6 +- 16)/2
= 3 +- 8
= -5 and +11.
Or you could just see that the factorization is
(t-11)(t+5) = 0.
t^2 = 6t + 55
t^2 - 6t - 55 = 0
(t + 5)(t - 11) = 0
t + 5 = 0
t = -5
t - 11 = 0
t = 11
â´ t = -5 , 11
收錄日期: 2021-05-01 10:15:16
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