How can the square root be eliminated in this algebraic manipulation problem?

[Nano-Passion]

Homework Statement

Simplify so that there is no square root function. Leave it in terms of S

S = xc^2 / (1 - 2Gy/rc^2)^1/2

s = \frac{xc^2}{(1-\frac{2Gy}{rc^2})^{1/2}}

Attempt

I've tried too many things, it would be too cumbersome to copy all of it down.

 

[rock.freak667]

Try squaring both sides and making the fraction simpler. Not sure which term you want as the subject though.

 

[Nano-Passion]

Try squaring both sides and making the fraction simpler. Not sure which term you want as the subject though.

I've done that, but ultimately we need it to be an explicit function of E, so you will still have to take the square root of the right side and it doesn't seem to be possible to simplify it.

 

[Mark44]

Homework Statement

Simplify so that there is no square root function. Leave it in terms of S

S = xc^2 / (1 - 2Gy/rc^2)^1/2

s = \frac{xc^2}{(1-\frac{2Gy}{rc^2})^{1/2}}

Attempt

I've tried too many things, it would be too cumbersome to copy all of it down.

You've been around here long enough to know the drill. You don't need to copy all of it down, but you do need to show some effort. I cut you some slack this time.

I've done that, but ultimately we need it to be an explicit function of E, so you will still have to take the square root of the right side and it doesn't seem to be possible to simplify it.

E? There's no E in this problem.

 

[Nano-Passion]

You've been around here long enough to know the drill. You don't need to copy all of it down, but you do need to show some effort. I cut you some slack this time.E? There's no E in this problem.

Sorry, mistype. I meant S.

You've been around here long enough to know the drill. You don't need to copy all of it down, but you do need to show some effort. I cut you some slack this time.E? There's no E in this problem.

Sorry about that.

 

[HallsofIvy]

What, exactly, do you mean by "no square root function"? You can, of course, immediately replace the square root by a 1/2 power, as you have done. Why is that not a correct answer? You can, of course, proceed to combine the fraction and simplify but you will still have 1/2 powers whatever you do.

 

[Nano-Passion]

What, exactly, do you mean by "no square root function"? You can, of course, immediately replace the square root by a 1/2 power, as you have done. Why is that not a correct answer? You can, of course, proceed to combine the fraction and simplify but you will still have 1/2 powers whatever you do.

Thanks for your time.

The 1/2 power is the equivalent of the square root so that defeats the purpose.

 

[Mark44]

Is the goal to get rid of the square root in the denominator?

 

[Nano-Passion]

Is the goal to get rid of the square root in the denominator?

Yes, but within the condition that equation is left as an explicit function of E.

 

[SOA Andrew]

Is the goal to get rid of the square root in the denominator?

Yes, but within the condition that equation is left as an explicit function of E.

Multiply the quotient on the right-hand side of the equation by a convenient form of 1; in this case, \frac{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}}{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}}

This technique is called rationalizing the denominator.

 

[SammyS]

Yes, but within the condition that equation is left as an explicit function of E.

There's that "E" again !

Perhaps ...

Are you supposed to be solving for y ?​

 

[Nano-Passion]

Multiply the quotient on the right-hand side of the equation by a convenient form of 1; in this case, \frac{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}}{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}}

This technique is called rationalizing the denominator.

s = \frac{xc^2}{(1-\frac{2Gy}{rc^2})^{1/2}} \frac{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}}{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}}

s = xc^2 \frac{\sqrt{1 - \frac{2Gy}{rc^2}}} {1-\frac{2Gy}{rc^2}}
s = xc^2 \sqrt{1 - \frac{2G}{rc^2}} (1 - \frac{rc^2}{2Gy})

There's that "E" again !

Perhaps ...

Are you supposed to be solving for y ?​

Whoops.

No, for S.

 

[SOA Andrew]

There is no longer a square root in the denominator, so you have succeeded in eliminating the square root in the denominator. However, I am afraid the last step is incorrect because

\frac{1}{1 - \frac{2Gy}{rc^2}} \neq 1 - \frac{rc^2}{2Gy}

The penultimate step is the last correct step.

 

[Nano-Passion]

There is no longer a square root in the denominator, so you have succeeded in eliminating the square root in the denominator. However, I am afraid the last step is incorrect because

\frac{1}{1 - \frac{2Gy}{rc^2}} \neq 1 - \frac{rc^2}{2Gy}

The penultimate step is the last correct step.

True, I've rewrote it below.

s = xc^2 \sqrt{1 - \frac{2G}{rc^2}} \frac{1}{1 - \frac{2Gy}{rc^2}}