Solve. t^2 + 3t - 28 = 0?
回答 (8)
2008-04-12 7:33 pm
✔ 最佳答案
t^2 + 3t - 28 = 0t^2+7t-4t-28=0
t(t+7)-4(t+7)=0
(t+7)(t-4)=0
t=4, t=-7
2008-04-15 7:30 pm
(t + 7)(t - 4) = 0
t = - 7 , t = 4
t = - 7 , t = 4
2008-04-13 2:41 am
t² + 3t - 28 = 0
(t + 7)(t - 4)
t + 7 = 0, t = - 7
t - 4 = 0, t = 4
Answer: t = - 7, 4
Proof (t is - 7):
7² + 3(- 7) = 28
49 - 21 = 28
Proof (t is 4):
4² + 3(4) = 28
16 + 12 = 28
(t + 7)(t - 4)
t + 7 = 0, t = - 7
t - 4 = 0, t = 4
Answer: t = - 7, 4
Proof (t is - 7):
7² + 3(- 7) = 28
49 - 21 = 28
Proof (t is 4):
4² + 3(4) = 28
16 + 12 = 28
2008-04-13 2:38 am
t²+3t - 28 = 0
Right off, I can see that t=4 and t=-7 are solutions. But if an answer doesn't jump out at you, the quadratic formula will work.
   at²+bt+c=0
where
   a=1
   b=3
   c=-28
t = [-b ± â(b²-4ac)] / (2a)
   = [-(3) ± â((3)² - 4(1)(-28))] / 2(1)
   = [-3 ± â(121)] / 2
   = [-3 ± 11] / 2
   = -1.5 ± 5.5
t₁ = -1.5 + 5.5 = 4
t₂ = -1.5 - 5.5 = -7
Right off, I can see that t=4 and t=-7 are solutions. But if an answer doesn't jump out at you, the quadratic formula will work.
   at²+bt+c=0
where
   a=1
   b=3
   c=-28
t = [-b ± â(b²-4ac)] / (2a)
   = [-(3) ± â((3)² - 4(1)(-28))] / 2(1)
   = [-3 ± â(121)] / 2
   = [-3 ± 11] / 2
   = -1.5 ± 5.5
t₁ = -1.5 + 5.5 = 4
t₂ = -1.5 - 5.5 = -7
2008-04-13 2:37 am
Hey, most of the people asking the same question and we are answer that we don't no weather you are getting or not so i will show you the method to do.
to find the value of x ,
ax^2 + bx + c = 0
would be factored into:
(mx+p)(nx+q) =0
Same time you can factorize and find the value of x.
where
mn = a,
pq = c, and
pn + mq = b.
I HOPE THIS WILL HELP!!!!!!!!!!!!!!
to find the value of x ,
ax^2 + bx + c = 0
would be factored into:
(mx+p)(nx+q) =0
Same time you can factorize and find the value of x.
where
mn = a,
pq = c, and
pn + mq = b.
I HOPE THIS WILL HELP!!!!!!!!!!!!!!
2008-04-13 2:35 am
t^2 + 3t - 28 = 0
(t + 7)(t - 4) = 0
t + 7 = 0
t = -7
t - 4 = 0
t = 4
â´ t = -7 , 4
(t + 7)(t - 4) = 0
t + 7 = 0
t = -7
t - 4 = 0
t = 4
â´ t = -7 , 4
2008-04-13 2:34 am
Factor:
(t-4)(t+7)=0
find (t-4=0) and (t+7=0):
t=4, -7
(t-4)(t+7)=0
find (t-4=0) and (t+7=0):
t=4, -7
2008-04-13 2:34 am
t^2 + 3t - 28 = 0
(t +7)(t - 4) = 0
t = 4, -7
(t +7)(t - 4) = 0
t = 4, -7
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