Subject of fomular

[Bekay]

Please i would like to make "Change T" subject of fomular. Please help.

 

[Dr.Peterson]

View attachment 24606
Please i would like to make "Change T" subject of fomular. Please help.

I would first isolate the square root (0.5 power), then square both sides. That should result in a quadratic equation you could solve by the quadratic formula.

Give it a try and show us your work, as we ask:

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[Bekay]

@ Dr.Peterson This is the extent i got to.

 

[Bekay]

I would first isolate the square root (0.5 power), then square both sides. That should result in a quadratic equation you could solve by the quadratic formula.

Give it a try and show us your work, as we ask:

Posting Guidelines (Summary)

Welcome to our tutoring boards! :) This page summarizes main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to help...

www.freemathhelp.com

@ Dr.Peterson This is the extent i got to. Please can you kindly help finish it up as it's quite urgent?

 

[Dr.Peterson]

First, "isolate" means to get it by itself on one side. To do that from what you have, add that 0.5 power to both sides to bring it to the left, and subtract IR/S. Then you'll have "square root of this = that"; when you square it, you'll get "this = that squared".

Second, I wouldn't do most of the work you did, expanding the square. In fact, I'd temporarily use u in place of \(T_H-\Delta T\) to keep things simple; or at least keep that expression intact throughout until near the end of my work.

Third, having now carried out the work, I see that the square will cancel out and you will be left with a linear equation to solve. No quadratic formula needed, unless I made a mistake.

 

[Bekay]

First, "isolate" means to get it by itself on one side. To do that from what you have, add that 0.5 power to both sides to bring it to the left, and subtract IR/S. Then you'll have "square root of this = that"; when you square it, you'll get "this = that squared".

Second, I wouldn't do most of the work you did, expanding the square. In fact, I'd temporarily use u in place of \(T_H-\Delta T\) to keep things simple; or at least keep that expression intact throughout until near the end of my work.

Third, having now carried out the work, I see that the square will cancel out and you will be left with a linear equation to solve. No quadratic formula needed, unless I made a mistake.

Many thanks for your response. I hope i am on the right track? The end goal is to make ΔT subject of fomular. What do you suggest i should do next?

 

[Dr.Peterson]

View attachment 24616

Many thanks for your response. I hope i am on the right track? The end goal is to make ΔT subject of fomular. What do you suggest i should do next?

At the last steps you didn't square the entire right-hand side; you left the negative unchanged! I would instead have moved the radical to the left and the rest to the right (or, equivalently, changed signs) before squaring.

But also, as I said, I would not expand the square under the radical. To avoid that temptation, as I suggested, let's replace `T-\Delta T` with u, and therefore `\Delta T` with `T-u`.

What I would have, at the step where you stopped, is `u^2-(2(T-u))/X = (u-(IR)/S)^2`. Now expand the square on the right and cancel the `u^2` on each side. Then solve for u, and finally subtract that from T to get `\Delta T`.

 

[Bekay]

At the last steps you didn't square the entire right-hand side; you left the negative unchanged! I would instead have moved the radical to the left and the rest to the right (or, equivalently, changed signs) before squaring.

But also, as I said, I would not expand the square under the radical. To avoid that temptation, as I suggested, let's replace `T-\Delta T` with u, and therefore `\Delta T` with `T-u`.

What I would have, at the step where you stopped, is `u^2-(2(T-u))/X = (u-(IR)/S)^2`. Now expand the square on the right and cancel the `u^2` on each side. Then solve for u, and finally subtract that from T to get `\Delta T`.

Wow! Thats great. Thanks so much