I know we have to first factor the polynomial on the left side - then I am lost.
2w(4w+1)=1
Using the zero factor property?
2008-12-05 2:43 pm
回答 (5)
2008-12-05 2:50 pm
✔ 最佳答案
You first have to undo the factor on the left by multiplying the 2w into the binomial:2w(4w + 1) = 1
8w² + 2w = 1
8w² + 2w - 1 = 0
Now factor again:
(4w - 1)(2w + 1) = 0
Now that you have tho product of two items equalling zero, you can use the zero factor properlty to set each item to zero and solve:
4w - 1 = 0 and 2w + 1 = 0
4w = 1 and 2w = -1
w = 1/4 and w = -1/2
2008-12-05 10:54 pm
The zero factor property says:
Let x be a variable, and {n₁, n₂, n₃ ... nₐ} be constants.
If
(x-n₁)(x-n₂)...(x-nₐ) = 0
Then
x-n₁=0
x-n₂=0
•
•
•
x-nₐ=0
Or something to that effect, so...
2w(4w+1) = 1
8w² + 2w = 1
8w² + 2w - 1 = 0
(4w - 1)(2w + 1) = 0
4w-1=0
4w=1
w = (1/4) <-- one answer
2w+1=0
2w = -1
w = -1/2 <-- other answer
Let x be a variable, and {n₁, n₂, n₃ ... nₐ} be constants.
If
(x-n₁)(x-n₂)...(x-nₐ) = 0
Then
x-n₁=0
x-n₂=0
•
•
•
x-nₐ=0
Or something to that effect, so...
2w(4w+1) = 1
8w² + 2w = 1
8w² + 2w - 1 = 0
(4w - 1)(2w + 1) = 0
4w-1=0
4w=1
w = (1/4) <-- one answer
2w+1=0
2w = -1
w = -1/2 <-- other answer
2008-12-05 10:54 pm
You'll need to get zero on the right, then factor:
LHS: 8w² + 2w = 1 (the RHS)
8w² + 2w - 1 = 0. NOW factor.
(2w+1)(4w-1) = 0 then apply the zero factor property to
find x= -1/2; or x= 1/4
LHS: 8w² + 2w = 1 (the RHS)
8w² + 2w - 1 = 0. NOW factor.
(2w+1)(4w-1) = 0 then apply the zero factor property to
find x= -1/2; or x= 1/4
2008-12-05 10:51 pm
2w(4w + 1) = 1
2w*4w + 2w*1 - 1 = 0
8w^2 + 2w - 1 = 0
8w^2 + 4w - 2w - 1 = 0
(8w^2 + 4w) - (2w + 1) = 0
4w(2w + 1) - 1(2w + 1) = 0
(2w + 1)(4w - 1) = 0
2w + 1 = 0
2w = -1
w = -1/2 (-0.5)
4w - 1 = 0
4w = 1
w = 1/4 (0.25)
â´ w = -1/2 (-0.5) , 1/4 (0.25)
2w*4w + 2w*1 - 1 = 0
8w^2 + 2w - 1 = 0
8w^2 + 4w - 2w - 1 = 0
(8w^2 + 4w) - (2w + 1) = 0
4w(2w + 1) - 1(2w + 1) = 0
(2w + 1)(4w - 1) = 0
2w + 1 = 0
2w = -1
w = -1/2 (-0.5)
4w - 1 = 0
4w = 1
w = 1/4 (0.25)
â´ w = -1/2 (-0.5) , 1/4 (0.25)
2008-12-05 10:49 pm
2w(4w+1) =1
8w^2 +2w -1 = 0
(2w+1)(4w-1) = 0
(2w+1) = 0 => w = -1/2
(4w-1) = 0 => w = 1/4
w1 = -1/2;
w2 = 1/4;
Bye
gilvi
8w^2 +2w -1 = 0
(2w+1)(4w-1) = 0
(2w+1) = 0 => w = -1/2
(4w-1) = 0 => w = 1/4
w1 = -1/2;
w2 = 1/4;
Bye
gilvi
收錄日期: 2021-05-01 11:35:52
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