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 :)

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

Quote:

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

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

Quote:

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

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

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

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

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

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

Quote:

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** $\displaystyle \sqrt{k}= \frac{2x}{x- 2}$.

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

Quote:

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!!

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?

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

Quote:

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 (Wink)

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

Quote:

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?

...

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?

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

Quote:

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?

...