Is this right or wrong???

2008-04-18 9:08 am
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??

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
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

∴ 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.
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.
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.
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}
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


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