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Algebra help!

How would I make x the subject of the formula when y=1/2(x+3) ?
Thanks
Reply 1
Original post by Toastiekid
How would I make x the subject of the formula when y=1/2(x+3) ?
Thanks


2y=x+3
2y-3=x
Reply 2
Original post by sz_rhl
2y=x+3
2y-3=x

Sorry i misread the question it's y=1/2(x+3)^2
I'm confused what you do about the quadratic
Reply 3
Original post by Toastiekid
Sorry i misread the question it's y=1/2(x+3)^2
I'm confused what you do about the quadratic


Hmmmm actually yeah that it quite difficult. I dropped maths last year along with my skills apparently. Sorry T__T
Reply 4
Original post by Toastiekid
Sorry i misread the question it's y=1/2(x+3)^2
I'm confused what you do about the quadratic


x(R{3}) x \in (\mathbb{R} - \{-3 \}) ( I guess so it should be specified ) then y>0 y >0 ... that's just the technical part.
Anyway it's not hard . Provided what I said (so that the thing under the square root is positive)
(x+3)2=12yx+3=±12yx=±12y3 (x+3)^2 = \frac{1}{2y} \Rightarrow x+3 =\pm \frac{1}{\sqrt{2y}} \Rightarrow x= \pm \frac{1}{\sqrt{2y}} -3
(edited 7 years ago)
The way you've written that could be written as (x+3)^2 all divided by 2.

So to start with I'd multiply y by 2 to get rid of the fraction on the right. So now we have 2y = (x+3)^2

Then square root the 2y to get rid of the power on the right. So now we have Square Root of 2y = x+3

Then it's simply subtract 3 from each side. So x = Square Root 2y - 3 (Remember the - 3 is NOT in the square root. It's separate)

I'm not 100% certain this is correct. I was very good at rearranging formulas in my GCSE and I don't ever remember doing anything special when it came to handling powers and roots etc. You don't need to work it out or anything as far as I know. You are just rearranging.

If you tried working out the quadratic on the right you end up with 2y = x^2 + 6x + 9 and then it gets really messy. I'm pretty sure it's what I wrote up top.
(edited 7 years ago)
Reply 6
Original post by Toastiekid
How would I make x the subject of the formula when y=1/2(x+3) ?
Thanks


First of all what do you mean
this y=12(x+3)2 y= \frac{1}{2} (x+3)^2 or y=12(x+3)2 y= \frac{1}{2(x+3)^2}
Edit: if it's the first everything is much simpler
2y=(x+3)2x+3=±2yx=±2y3 2y=(x+3)^2 \Rightarrow x+3=\pm \sqrt{2y} \Rightarrow x= \pm \sqrt{2y} - 3
(edited 7 years ago)
Reply 7
@I'm Your Teddy Bear Did you tag me in a comment that you deleted ?
Edit: Also are you from Iran ?
(edited 7 years ago)
Reply 8
Original post by Darwinion
The way you've written that could be written as (x+3)^2 all divided by 2.

So to start with I'd multiply y by 2 to get rid of the fraction on the right. So now we have 2y = (x+3)^2

Then square root the 2y to get rid of the power on the right. So now we have Square Root of 2y = x+3

Then it's simply subtract 3 from each side. So x = Square Root 2y - 3 (Remember the - 3 is NOT in the square root. It's separate)

I'm not 100% certain this is correct. I was very good at rearranging formulas in my GCSE and I don't ever remember doing anything special when it came to handling powers and roots etc. You don't need to work it out or anything as far as I know. You are just rearranging.

If you tried working out the quadratic on the right you end up with 2y = x^2 + 6x + 9 and then it gets really messy. I'm pretty sure it's what I wrote up top.


Thank you! I checked with my teacher and that was the working they did (the first one) by making it quadratic I was making it harder than it needed to be!
Reply 9
Original post by Vesniep
First of all what do you mean
this y=12(x+3)2 y= \frac{1}{2} (x+3)^2 or y=12(x+3)2 y= \frac{1}{2(x+3)^2}
Edit: if it's the first everything is much simpler
2y=(x+3)2x+3=±2yx=±2y3 2y=(x+3)^2 \Rightarrow x+3=\pm \sqrt{2y} \Rightarrow x= \pm \sqrt{2y} - 3

It's the first- sorry for how it's written I just copied the question from the test paper. Thank you!

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