Thread: finding x when it appears in the equation more than once

Hi, this is the first time I have come across a question where x has appeared in an equation more than once so I am a bit stumped as how to find it.

55.17 = [(2x)^2]/[(2-x)(2-x)]
x = ?

I can simply it to:
55.17 = (4x^2) / [(2-x)^2]
55.17^(1/2) = 4x / (2-x)
but still can't find x

any help would be great

Originally Posted by Siddy

Hi, this is the first time I have come across a question where x has appeared in an equation more than once so I am a bit stumped as how to find it.

55.17 = [(2x)^2]/[(2-x)(2-x)]
x = ?

I can simply it to:
55.17 = (4x^2) / [(2-x)^2]
55.17^(1/2) = 4x / (2-x)
but still can't find x

any help would be great

denote .

Then

55.17^(1/2) = 4x / (2-x)

the OP is wrong here,it should be 2x/(2-x)

Thanks so much abhishekkgp,
can you tell me how you did this bit:

k(2 - x) = 4x => 2k - 2x = 4x ?

I thought it would be:

k(2 - x) = 4x => 2k - kx = 4x

Thanks anonimnystefy, i forgot about square rooting the 2 there.

Originally Posted by Siddy

Thanks so much abhishekkgp,
can you tell me how you did this bit:

k(2 - x) = 4x => 2k - 2x = 4x ?

I thought it would be:

k(2 - x) = 4x => 2k - kx = 4x

Yes, you are right. That must have been a typo.

However, you are not done. from you can get both and .

Originally Posted by Siddy

Thanks so much abhishekkgp,
can you tell me how you did this bit:

k(2 - x) = 4x => 2k - 2x = 4x ?

I thought it would be:

k(2 - x) = 4x => 2k - kx = 4x

my mistake!!the correct thing is k(2-x)=2k-kx.
SORRY!!

Thanks HallsofIvy,

solution 1.

K^1/2 = 2x / 2 - x
K^1/2 (2 - x) = 2x
[K^1/2 (2 - x )]/2 = x
(2.K^1/2 - x.K^1/2)/2 = x
(2.K^1/2)/2 = x + x.K^1/2
(2.K^1/2)/2.K^1/2 = 2x
(2.K^1/2)/4.K^1/2 = x
1/2 = x


solution 2.
K^1/2 = 2x / x - 2
K^1/2 (x - 2) = 2x
x.K^1/2 - 2K^1/2 = 2x
-2K^1/2 = 2x - x.K^1/2
(-2K^1/2) / K^1/2 = 2x - x
-K^1/2 = x

The answer can not be negative therefore solution 1 must be correct (as long as my working out is right).

EDIT: actually i think i must have done something wrong because the answer should utilize K, but i cancelled it out. Because the answer needs to be different for different values of K.
Can you guys point out my error?

Originally Posted by abhishekkgp

my mistake!!the correct thing is k(2-x)=2k-kx.
SORRY!!

No worries abhishekkgp, you still got me going in the right direction

Originally Posted by Siddy

Thanks HallsofIvy,

solution 1.

K^1/2 = 2x / 2 - x
K^1/2 (2 - x) = 2x
[K^1/2 (2 - x )]/2 = x
(2.K^1/2 - x.K^1/2)/2 = x
(2.K^1/2)/2 = x + x.K^1/2 shouldn't this be x+(1/2)(x.sqrt(k))??
(2.K^1/2)/2.K^1/2 = 2x
(2.K^1/2)/4.K^1/2 = x
1/2 = x


solution 2.
K^1/2 = 2x / x - 2
K^1/2 (x - 2) = 2x
x.K^1/2 - 2K^1/2 = 2x
-2K^1/2 = 2x - x.K^1/2
(-2K^1/2) / K^1/2 = 2x - x
-K^1/2 = x

The answer can not be negative therefore solution 1 must be correct (as long as my working out is right).

EDIT: actually i think i must have done something wrong because the answer should utilize K, but i cancelled it out. Because the answer needs to be different for different values of K.
Can you guys point out my error?

...

Thanks again abhishekkgp,

solution 1.

K^1/2 = 2x / 2 - x
K^1/2 (2 - x) = 2x
[K^1/2 (2 - x )]/2 = x
(2.K^1/2 - x.K^1/2)/2 = x
(2.K^1/2)/2 = x + (x.K^1/2)/2
(4.K^1/2)/2 = 2x + x.K^1/2 <- but I still get stuck and can't separate the x's and the K's...


I tried it again and went this way:
K^1/2 = 2x / 2 - x
K^1/2 (2 - x) = 2x
K^1/2 (2 - x ) = 2x
2.K^1/2 - x.K^1/2 = 2x
2 - x = 2x / K^1/2
2 = (2x + x) / K^1/2
2.K^1/2 = 3x
(2.K^1/2) / 3 = x

Could that be right?

Also, if it is right, how come I am stuck in the first part in this post, is there a trick I am missing?

Originally Posted by Siddy

Thanks again abhishekkgp,

solution 1.

K^1/2 = 2x / 2 - x
K^1/2 (2 - x) = 2x
[K^1/2 (2 - x )]/2 = x
(2.K^1/2 - x.K^1/2)/2 = x
(2.K^1/2)/2 = x + (x.K^1/2)/2
(4.K^1/2)/2 = 2x + x.K^1/2 <- but I still get stuck and can't separate the x's and the K's...
this gives 2.sqrt(k)=x(2+sqrt(k)) which gives x=(2.sqrt(k))/(2+sqrt(k))


I tried it again and went this way:
K^1/2 = 2x / 2 - x
K^1/2 (2 - x) = 2x
K^1/2 (2 - x ) = 2x
2.K^1/2 - x.K^1/2 = 2x
2 - x = 2x / K^1/2
2 = (2x + x) / K^1/2
2.K^1/2 = 3x
(2.K^1/2) / 3 = x

Could that be right?

Also, if it is right, how come I am stuck in the first part in this post, is there a trick I am missing?

...