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

1. ## 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

2. ## Re: finding x when it appears in the equation more than once

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 $\displaystyle \sqrt{55.17}=k$.

Then $\displaystyle k= \frac{4x}{2-x} \Rightarrow k(2-x)=4x \Rightarrow 2k-2x=4x \Rightarrow 2k=4x+2x \Rightarrow 2k=6x \Rightarrow k/3=x$

3. ## Re: finding x when it appears in the equation more than once

55.17^(1/2) = 4x / (2-x)
the OP is wrong here,it should be 2x/(2-x)

4. ## Re: finding x when it appears in the equation more than once

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

5. ## Re: finding x when it appears in the equation more than once

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

6. ## Re: finding x when it appears in the equation more than once

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 $\displaystyle k= \frac{4x^2}{(2- x)^2}$ you can get both $\displaystyle \sqrt{k}= \frac{2x}{2- x}$ and $\displaystyle \sqrt{k}= \frac{2x}{x- 2}$.

7. ## Re: finding x when it appears in the equation more than once

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!!

8. ## Re: finding x when it appears in the equation more than once

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?

9. ## Re: finding x when it appears in the equation more than once

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

10. ## Re: finding x when it appears in the equation more than once

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?
...

11. ## Re: finding x when it appears in the equation more than once

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?

12. ## Re: finding x when it appears in the equation more than once

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?
...