Make ##x## the subject of the formula

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In summary, this was a good one; just thought i would share...of course i know this forum has smart people who may probably come up with a different approach...it took me a few minutes to figure out ...:frown:and then alas! saw the magic...
  • #1

chwala

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Homework Statement
Make ##x## the subject of formula, given;

##\sqrt{\dfrac {x^2+15t-3}{x^2}}=5t##
Relevant Equations
Basic Maths
This was a good one; just thought i would share...of course i know this forum has smart people who may probably come up with a different approach...it took me a few minutes to figure out ...:frown:and then alas! saw the magic...

##\sqrt{\dfrac {x^2+15t-3}{x^2}}=5t##

...

##x(5t+1)(5t-1)=15t-3##

##x=\dfrac{3}{5t+1}##
Bingo!
 
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  • #2
chwala said:
Homework Statement:: Make ##x## the subject of formula, given;

##\dfrac{\sqrt{x^2+15t-3}}{x^2}=5t##
Relevant Equations:: Basic Maths

This was a good one; just thought i would share...of course i know this forum has smart people who may probably come up with a different approach...it took me a few minutes to figure out ...:frown:and then alas! saw the magic...

##\sqrt{\dfrac {x^2+15t-3}{x^2}}=5t##

...

##x(5t+1)(5t-1)=15t-3##

##x=\dfrac{3}{5t+1}##
Bingo!
Are you sure that’s correct? It’s always worth back-checking. For example,

Let ##t = 0##, then ##x = \dfrac 3{5t+1} = \dfrac 3{5*0+1} = 3##

##\dfrac {\sqrt{x^2+15t-3}}{x^2}
=\dfrac {\sqrt{3^2+15*0-3}}{3^2}
=\dfrac {\sqrt 6}{9}##

But ##5t=5*0 = 0##
 
  • #3
Steve4Physics said:
Are you sure that’s correct? It’s always worth back-checking. For example,

Let ##t = 0##, then ##x = \dfrac 3{5t+1} = \dfrac 3{5*0+1} = 3##

##\dfrac {\sqrt{x^2+15t-3}}{x^2}
=\dfrac {\sqrt{3^2+15*0-3}}{3^2}
=\dfrac {\sqrt 6}{9}##

But ##5t=5*0 = 0##
I'll check this later man! Sorry I made an amendment on the problem...latex problem...

My working ought to be correct...sorry currently in a meeting...hence delay in responding...
 
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  • #4
chwala said:
Homework Statement:: Make ##x## the subject of formula, given;

##\dfrac{\sqrt{x^2+15t-3}}{x^2}=5t##
Relevant Equations:: Basic Maths

This was a good one; just thought i would share...of course i know this forum has smart people who may probably come up with a different approach...it took me a few minutes to figure out ...:frown:and then alas! saw the magic...

##\sqrt{\dfrac {x^2+15t-3}{x^2}}=5t##

This is not the same as the equation you give in the problem statement. Does the square root apply to the whole fraction, or only the numerator?
 
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  • #5
pasmith said:
This is not the same as the equation you give in the problem statement. Does the square root apply to the whole fraction, or only the numerator?
Amended :wink:.
 
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  • #6
chwala said:
I'll check this later man! Sorry I made an amendment on the problem...latex problem...

My working ought to be correct...sorry currently in a meeting...hence delay in responding...
Your working of the problem still appears to be incorrect.

The counter-example given by @Steve4Physics remains valid and shows that your solution is incorrect.
 
  • #7
SammyS said:
Your working of the problem still appears to be far from correct.

The counter-example given by @Steve4Physics remains valid and shows that your solution is incorrect.
...I have seen the mistake...let me correct it later...should be square root on rhs...
 
  • #8
chwala said:
Homework Statement:: Make ##x## the subject of formula, given;

##\sqrt{\dfrac {x^2+15t-3}{x^2}}=5t##
Relevant Equations:: Basic Maths

This was a good one; just thought i would share...of course i know this forum has smart people who may probably come up with a different approach...it took me a few minutes to figure out ...:frown:and then alas! saw the magic...

##\sqrt{\dfrac {x^2+15t-3}{x^2}}=5t##

...

##x^2(5t+1)(5t-1)=15t-3##

##x=\sqrt{\dfrac{3}{5t+1}}##
Bingo!
 
  • #9
You're nearly there...

1) You are missing a ± sign.

2) It would be good practice to include a statement about the allowed values of t.

3) The word 'alas' is used when something has gone badly - not when it has gone well!
 
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  • #10
chwala said:
Homework Statement:: Make ##x## the subject of formula, given;

##\sqrt{\dfrac {x^2+15t-3}{x^2}}=5t##
Relevant Equations:: Basic Maths

This was a good one; just thought i would share...of course i know this forum has smart people who may probably come up with a different approach...it took me a few minutes to figure out ...:frown:and then alas! saw the magic...

##\sqrt{\dfrac {x^2+15t-3}{x^2}}=5t##

...

##x(5t+1)(5t-1)=15t-3##

##x=\dfrac{3}{5t+1}##
Yes, I am aware that you amended the last two lines of your solution in Post #8. (However, your post is done totally as a Quote of the OP.) You had also amended the original OP, as you mentioned in Posts #3 & #5.

The two last lines as you amended them:

##\displaystyle x^2(5t+1)(5t-1)=15t-3##​
##\displaystyle x=\sqrt{\dfrac{3}{5t+1}} ##​

Comments:

It would have been at least helpful if you had included a couple of steps showing what you did to get to those last two lines, rather than simply giving us "##\ \dots\ ##" . That would have made it obvious that that the Homework Statement was in error. Of course that would have made your solution less magical.

Similarly, but in my opinion even more importantly, you should have included some steps missing between those final two equations, as well as justification.

As I recall, you have been cautioned about cancelling a factor by division, when that factor contains a variable.
 
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  • #11
SammyS said:
Yes, I am aware that you amended the last two lines of your solution in Post #8. (However, your post is done totally as a Quote of the OP.) You had also amended the original OP, as you mentioned in Posts #3 & #5.

The two last lines as you amended them:

##\displaystyle x^2(5t+1)(5t-1)=15t-3##​
##\displaystyle x=\sqrt{\dfrac{3}{5t+1}} ##​

Comments:

It would have been at least helpful if you had included a couple of steps showing what you did to get to those last two lines, rather than simply giving us "##\ \dots\ ##" . That would have made it obvious that that the Homework Statement was in error. Of course that would have made your solution less magical.

Similarly, but in my opinion even more importantly, you should have included some steps missing between those final two equations, as well as justification.

As I recall, you have been cautioned about cancelling a factor by division, when that factor contains a variable.
Noted @SammyS ... let me re do it again; here we go##\sqrt{\dfrac {x^2+15t-3}{x^2}}=5t##

##\dfrac{x^2+15t-3}{x^2}=25t^2##

##x^2+15t-3=25x^2t^2##

##15t-3=25x^2t^2-x^2##

##15t-3=x^2(25t^2-1)##

...Actually, there was no need for me to have used ##a^2-b^2=(a-b)(a+b)## previously.

##x^2=\dfrac{15t-3}{25t^2-1}##

##x=±\sqrt{\dfrac{15t-3}{25t^2-1}}##

...i guess what i pick from you is the not to cancel variable bit...and also the plus & minus as indicated by @Steve4Physics ...thanks for that...
 
  • #12
Steve4Physics said:
You're nearly there...

1) You are missing a ± sign.

2) It would be good practice to include a statement about the allowed values of t.

3) The word 'alas' is used when something has gone badly - not when it has gone well!
This question is from a singaporean paper...i believe for the early years (secondary)... the question is posted as it is on the paper. They did not ask the students to state the domain of the function for which the inverse function exists ( ##t## values as you indicate)... we know that the inverse function will only exist if we restrict the domain to i.e 1-1 relationship between the function and its inverse...

...to answer your question by looking at the two graphs, the inverse exists when the domain is restricted to ##x≥0##.
 
  • #13
chwala said:
##15t-3=x^2(25t^2-1)##

...Actually, there was no need for me to have used ##a^2-b^2=(a-b)(a+b)## previously.

##x^2=\dfrac{15t-3}{25t^2-1}##

##x=±\sqrt{\dfrac{15t-3}{25t^2-1}}##

...i guess what i pick from you is the not to cancel variable bit...and also the plus & minus as indicated by @Steve4Physics ...thanks for that...
Thanks for taking us to ## \displaystyle 15t-3=x^2(25t^2-1)## .

Factoring the difference of squares is a great idea. After also factoring the LHS, that gets us to
## \displaystyle 3(5t-1)=x^2(5t-1)(5t+1)## .

You may divide both sides by ## \displaystyle (5t-1)## provided it's not zero, so we need to handle ##\displaystyle t=\dfrac{1}{5}## in some other way.

A better way to handle this in general, is to get zero on one side of an equivalent equation and then factor if you can. In this case we get the following.

## \displaystyle 0=x^2(5t-1)(5t+1) - 3(5t-1)##

## \displaystyle 0=(x^2(5t+1) - 3)(5t-1)##

So either ## \displaystyle x^2(5t+1) - 3 = 0 ## or ## \displaystyle 5t-1 = 0## .
 
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  • #14
SammyS said:
So either ## \displaystyle x^2(5t+1) - 3 = 0 ## or ## \displaystyle 5t-1 = 0## .
If ## \displaystyle x^2(5t+1) - 3 = 0 ##, then you have the solution given in Post #11.

What to do with ## \displaystyle 5t-1 = 0## ?

Plugging ## \displaystyle t = \dfrac 1 5 ## into ##\displaystyle \sqrt{\dfrac {x^2+15t-3}{x^2}}=5t ## gives:

##\displaystyle \sqrt{\dfrac {x^2}{x^2}}=1 ##.

This gives you that ##x## may be any non-zero real number when ## \displaystyle t = \dfrac 1 5 ## .
 
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  • #15
chwala said:
##x(5t+1)(5t-1)=15t-3##
##x^2(5t+1)(5t-1)=15t-3##
 
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  • #16
neilparker62 said:
##x^2(5t+1)(5t-1)=15t-3##
Thanks...missed that on latex...
 

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