2u^2=-7u-5
This is how I worked it out:
4u=-7u-5
so I added 7 to both sides and got:
11u=-5...divided by 11
So answer is u=-5/11
YES OR NO??
Is this right or wrong???
2008-04-18 9:08 am
回答 (9)
2008-04-18 9:13 am
✔ 最佳答案
No!4u = 2^2*u
Here goes:
Transpose the left side to the right
2u^2 +7u +5 = 0
Factor
(2u+5)(u+1) = 0
u = -5/2
u = -1
2008-04-18 4:28 pm
2u² = - 7u - 5
2u² + 7u + 5 = 0
(2u + 5)(u + 1) = 0
2u + 5 = 0, 2u = - 5, u = - 5/2
u + 1 = 0, u = - 1
Answer: u = - 5/2, - 1
Check (u = - 5/2):
2([-5/2]²) = - 7(- 5/2) - 5
2(25/4) = 35/2 - 10/2
25/2 = 25/2
Check (u = - 1):
2(- 1²) = - 7(- 1) - 5
2(1) = 7 - 5
2 = 2
2u² + 7u + 5 = 0
(2u + 5)(u + 1) = 0
2u + 5 = 0, 2u = - 5, u = - 5/2
u + 1 = 0, u = - 1
Answer: u = - 5/2, - 1
Check (u = - 5/2):
2([-5/2]²) = - 7(- 5/2) - 5
2(25/4) = 35/2 - 10/2
25/2 = 25/2
Check (u = - 1):
2(- 1²) = - 7(- 1) - 5
2(1) = 7 - 5
2 = 2
2008-04-18 5:31 pm
2u^2 = -7u - 5
2u^2 + 7u + 5 = 0
(2u + 5)(u + 1) = 0
2u + 5 = 0
2u = -5
u = -5/2
u = -2 1/2 (-2.5)
u + 1 = 0
u = -1
â´ u = -2 1/2 (-2.5) , -1
2u^2 + 7u + 5 = 0
(2u + 5)(u + 1) = 0
2u + 5 = 0
2u = -5
u = -5/2
u = -2 1/2 (-2.5)
u + 1 = 0
u = -1
â´ u = -2 1/2 (-2.5) , -1
2008-04-18 4:21 pm
No. that is not correct.
rearrange the terms to:
2u^2 + 7u + 5 = 0
then you can factor to:
(2u + 5) (u + 1) = 0
the only way two numbers multiplied together equals zero is if one of the numbers is zero.
So either
2u + 5 = 0
or
u + 1 = 0
Either of which is easily solved. This problem has two answers.
rearrange the terms to:
2u^2 + 7u + 5 = 0
then you can factor to:
(2u + 5) (u + 1) = 0
the only way two numbers multiplied together equals zero is if one of the numbers is zero.
So either
2u + 5 = 0
or
u + 1 = 0
Either of which is easily solved. This problem has two answers.
2008-04-18 4:16 pm
hey , to check whether your answer is correct or not , just put the value of variable( here u) in the equation. If you get LHS=RHS, your answer is correct.
2008-04-18 4:15 pm
NO
2u^2 does *not* equal 4u. Take this example. Let's say u=3, so 2(3)^2=2*9=18, but 4(3) = 12. Clearly those two are not the same.
The correct way to do it is to add 7u + 5 to both sides to get 2u^2+7u+5=0. From there, you factor it. *shudder* I can't quite explain it without giving away the answer (although I'm sure somebody else will.) But keep plugging in numbers into (ax+b)(cx+d) where a,b,c, and d are some numbers until it works.
2u^2 does *not* equal 4u. Take this example. Let's say u=3, so 2(3)^2=2*9=18, but 4(3) = 12. Clearly those two are not the same.
The correct way to do it is to add 7u + 5 to both sides to get 2u^2+7u+5=0. From there, you factor it. *shudder* I can't quite explain it without giving away the answer (although I'm sure somebody else will.) But keep plugging in numbers into (ax+b)(cx+d) where a,b,c, and d are some numbers until it works.
2008-04-18 4:14 pm
I see you have made 2u^2 into 4u. This is incorrect, the only way that it would be correct is if it were in brackets. So (2u)^2 which would then make it 4u^2.
Then put all the numbers on one side so it equals 0. Then just factorize it.
Then put all the numbers on one side so it equals 0. Then just factorize it.
2008-04-18 4:14 pm
2u² = -7u - 5
2u² + 7u + 5 = 0
2u² + 2u + 5u + 5 = 0
2u(u+1) + 5(u+1) = 0
(2u+5)(u+1) = 0
u = {-5/2, -1}
2u² + 7u + 5 = 0
2u² + 2u + 5u + 5 = 0
2u(u+1) + 5(u+1) = 0
(2u+5)(u+1) = 0
u = {-5/2, -1}
2008-04-18 4:16 pm
no
2 times u squared is not the same as 2 squared times u
you add 7u and 5 to both sides to get:
2u^2 + 7u +5 = 0
and factor by grouping - you can re-write 7u as 2u+5u so
2u^2 + 2u + 5u+ 5 = 0
and factor the bits
2u(u+1) + 5(u+1) = 0
then factor out the (u+1)
(u+1)(2u+5)=0
so u+1=0 or 2u+5=0
so u=-1 or u=-5/2
2 times u squared is not the same as 2 squared times u
you add 7u and 5 to both sides to get:
2u^2 + 7u +5 = 0
and factor by grouping - you can re-write 7u as 2u+5u so
2u^2 + 2u + 5u+ 5 = 0
and factor the bits
2u(u+1) + 5(u+1) = 0
then factor out the (u+1)
(u+1)(2u+5)=0
so u+1=0 or 2u+5=0
so u=-1 or u=-5/2
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