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# Simple equation rearrangement

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1. Hi,

How would you rearrange this equation so you end up with the quadratic 1/2p2 + p - 12 = 0. The equation is: 50p = 100 (12/p - 1).

I have tried everything but I keep getting 50p + 100p -1200 = 0
2. (Original post by Logolept)
Hi,

How would you rearrange this equation so you end up with the quadratic 1/2p2 + p + 12 = 0. The equation is: 50p = 100 (12/p - 1).

I have tried everything but I keep getting 50p + 100p -1200 = 0
Can you show your working that you've done?
3. The final step would be to divide everything by 100, but that wouldn't give the same as the top equation so you must have typed the last one in wrong (swapped a plus an minus sign?).
4. (Original post by Slowbro93)
Can you show your working that you've done?
Alright but please don't laugh at me lol.

Here:

50p = 100 (12/ p - 1)
50p = 1200 / p - 100
50p * p = 1200 - 100p
50p^2 + 100p - 1200 = 0
5. (Original post by B_9710)
The final step would be to divide everything by 100, but that wouldn't give the same as the top equation so you must have typed the last one in wrong (swapped a plus an minus sign?).
why would you divide by 100 though? Also yes, it is meant to be a minus sign. My mistake.
6. (Original post by Logolept)
why would you divide by 100 though?
if you have two sides of an equation, you can do the same thing to both of them and the equation is still true, for example:

.

I can divide both sides by 10 to get .

So, let's extend this to less obvious things.

If we have , we can divide both sides by 2 to get:

.

So, if you have quadratic like , you can divide both sides by a 100 to get:

7. (Original post by Zacken)
if you have two sides of an equation, you can do the same thing to both of them and the equation is still true, for example:

.

I can divide both sides by 10 to get .

So, let's extend this to less obvious things.

If we have , we can divide both sides by 2 to get:

.

So, if you have quadratic like , you can divide both sides by a 100 to get:

I understand that, but why 100, isn't the LCF here 50?
8. (Original post by Logolept)
I understand that, but why 100, isn't the LCF here 50?
To make the coefficient of x2 term 1/2 like the equation at the top of your post.
9. (Original post by Logolept)
I understand that, but why 100, isn't the LCF here 50?
Because you want to show that you can write the quadratic as "1/2 p^2 + ..."

And you have , so the obvious thing to do is to divide by , it'll give you a fractional coefficient which is normally ugly, but the question is asking for it.
10. Maybe I can put this into some context. Here is the original question (part c). In the answers I have been made aware that we must reach that quadratic equation which we put into the quadratic formula.
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11. (Original post by Zacken)
Because you want to show that you can write the quadratic as "1/2 p^2 + ..."

And you have , so the obvious thing to do is to divide by , it'll give you a fractional coefficient which is normally ugly, but the question is asking for it.
Sorry about the confusion. The question does not actually ask me to do that. I have posted the question in my new post.
12. (Original post by Logolept)
Maybe I can put this into some context. Here is the original question (part c). In the answers I have been made aware that we must reach that quadratic equation which we put into the quadratic formula.
What you need to understand that any of the quadratic equations you arrive at are correct, they will all give you the same answer for .

So if you want to solve by dividing by 50 instead, that's more than fine and it's completely equivalent. You will get the same answer. Dividing both sides of the equation by any number will not change the answer.
13. (Original post by Zacken)
What you need to understand that any of the quadratic equations you arrive at are correct, they will all give you the same answer for .

So if you want to solve by dividing by 50 instead, that's more than fine and it's completely equivalent. You will get the same answer. Dividing both sides of the equation by any number will not change the answer.
Got you! Thank you for being patient with me.
14. (Original post by Logolept)
Got you! Thank you for being patient with me.
No worries, glad you're asking questions on TSR, it shows a real drive for learning and understanding and it's admirable of you.

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