Can somebody help: 4t^2 = -6t-2?

2008-08-17 6:31 pm
4t^2 = -6t-2

^ = exponent

Can somebody explain how to solve this and give me the answer as well?

回答 (5)

2008-08-17 6:43 pm
✔ 最佳答案
2 t ² + 3 t + 1 = 0
( 2 t + 1 ) ( t + 1 ) = 0
t = - 1/2 , t = - 1
2008-08-18 1:59 am
Add 6t to both sides

4t^2 + 6t = -6t + 6t - 2 = -2

Add 2 to both sides

4t^2 + 6t + 2 = -2 + 2 = 0

Divide through by 2

2t^2 + 3t + 1 = 0

This factors as

(2t + 1)(t + 1) = 0

2t + 1 = 0 -> 2t = -1 -> t = -1/2

t + 1 = 0 -> t = -1

參考: Longtime college math teacher
2008-08-18 1:51 am
4t^2 = -6t - 2
4t^2 + 6t + 2 = 0
(4t^2 + 6t + 2)/2 = 0/2
2t^2 + 3t + 1 = 0
2t^2 + 2t + t + 1 = 0
(2t^2 + 2t) + (t + 1) = 0
2t(t + 1) + 1(t + 1) = 0
(t + 1)(2t + 1) = 0

t + 1 = 0
t = -1

2t + 1 = 0
2t = -1
t = -1/2 (-0.5)

∴ t = -1 , -1/2 (-0.5)
2008-08-18 1:41 am
Sure. Let's set the equation equal to zero: 4t^2 + 6t + 2 = 0. We know this has to factor into (_t + _) * (_t + _), cause everything is positive. So we need two numbers that, when multiplied together, give us 2. Well, that's got to be 1 and 2. Let's put them in: (_t + 2) * (_t + 1). Now, we need two numbers that, when multiplied together, give us 4. That could be 2 and 2 or 1 and 4. But, when you multiply the t terms together, it has to equal 6t. The only way that can happen is if we put the 4 in the first term and the 1 in the second term: (t + 1) * (4t + 2) = 4t^2 + 6t + 2.

So, since (t + 1) * (4t + 2) = 0, one of those two terms has to equal 0. If t + 1 = 0, then t = -1. If 4t + 2 = 0, then t = -1/2. So your answer is t = {-1, -1/2}.

If that wasn't clear, google "FOIL method". You'll get plenty of results.
2008-08-18 1:35 am
4t^2 + 6t + 2 = 0
(4t+2)(t+1)=0

4t+2 = 0
t = -1/2

t+1 = 0
t = -1


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