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# If x is positive, why take the negative root? watch

1. I'm confused by this. The question asks to find coordinates of A where both x and y are positive.

It's a simultaneous equation where the answer is
x = +/- 2(root 3) + 4

Yet, the mark scheme used this?:

It turns out that either way, whether you'd use the positive, or the negative root, x would be positive. Why did they choose to use the -2(root3) + 4? I thought that the larger value would surely be more appropriate?
2. (Original post by frostyy)
...
From the sketch. What does the sketch look like? Does it show that x is less than 4 in anyway?
3. Forgive me if i'm wrong, but the question wants where the coordinates are both positive. If you use 4+2(root3), your value for y is negative. Therefore, it doesn't fit what the question is asking you for?
4. (Original post by Zacken)
From the sketch. What does the sketch look like? Does it show that x is less than 4 in anyway?
If you mean the sketches from the previous part, x was actually equal to 0, 4 and 7.
5. (Original post by frostyy)
If you mean the sketches from the previous part, x was actually equal to 0, 4 and 7.
Could you show us the whole question, if possible?
6. (Original post by backtraced)
Forgive me if i'm wrong, but the question wants where the coordinates are both positive. If you use 4+2(root3), your value for y is negative. Therefore, it doesn't fit what the question is asking you for?
That probably is the explanation. Thanks!
7. (Original post by Zacken)
Could you show us the whole question, if possible?

This is the question. By the way, could you explain to me by any chance why this attempt in solving the simultaneous equation for c) didn't work?
8. (Original post by frostyy)
That probably is the explanation. Thanks!
You're welcome! Does it make sense? If not, I can find the question you're doing and try and draw a graph etc to explain it/prove it for you if you'd like. I'm sorry if my post seemed really blunt!
9. (Original post by frostyy)
...
Have you posted the wrong picture?
10. (Original post by frostyy)
This is the question. By the way, could you explain to me by any chance why this attempt in solving the simultaneous equation for c) didn't work?
What you've done in the picture isn't quite correct. Instead of cancelling out the x's from both sides you should have brought everything to one side factorised. Otherwise you lost the solution .

Anyhow, even if you did cancel, you should end up with

So that you have .
11. (Original post by Zacken)
Have you posted the wrong picture?
Forgive me, it's a bit late ;p.

12. (Original post by frostyy)
This is the question. By the way, could you explain to me by any chance why this attempt in solving the simultaneous equation for c) didn't work?
I think you've posted the wrong question. I'm assuming its the same question as in the original post though.

I think your issue is where you've divided through by x. You've over complicated it, and that would be the same as dividing through by x^2, not x I believe.

Honestly, the way i'd do it is like this:

x(4-x) = x^2(7-x)
DIVIDE BY X
(4-x) = x(7-x)
EXPAND
4-x=7x-x^2
REARRANGE SO x^2 IS POSITIVE AND IT EQUALS ZERO
x^2 - 8x + 4 = 0.
Use the quadratic formula, and it gives the answer from the markscheme you originally posted.
13. (Original post by backtraced)
I think you've posted the wrong question. I'm assuming its the same question as in the original post though.

I think your issue is where you've divided through by x. You've over complicated it, and that would be the same as dividing through by x^2, not x I believe.

Honestly, the way i'd do it is like this:

x(4-x) = x^2(7-x)
DIVIDE BY X
(4-x) = x(7-x)
EXPAND
4-x=7x-x^2
REARRANGE SO x^2 IS POSITIVE AND IT EQUALS ZERO
x^2 - 8x + 4 = 0.
Use the quadratic formula, and it gives the answer from the markscheme you originally posted.

The bit I've bolded. Don't do this.

Move everything over to the one side instead and then factorise an out.
14. (Original post by backtraced)
I think you've posted the wrong question. I'm assuming its the same question as in the original post though.

I think your issue is where you've divided through by x. You've over complicated it, and that would be the same as dividing through by x^2, not x I believe.

Honestly, the way i'd do it is like this:

x(4-x) = x^2(7-x)
DIVIDE BY X
(4-x) = x(7-x)
EXPAND
4-x=7x-x^2
REARRANGE SO x^2 IS POSITIVE AND IT EQUALS ZERO
x^2 - 8x + 4 = 0.
Use the quadratic formula, and it gives the answer from the markscheme you originally posted.
Thank you! Just to be clear on this though, I thought that if I have x outside the brackets on one side of the equation, I should divide everything on the other side by it to take it out?

So x(4-x) = x^2(7-x)
4 - x = (x^2 / x)((7 - x) / x)

I understand that I'm wrong now though.
15. (Original post by frostyy)
Forgive me, it's a bit late ;p.
Your sketch is like this:

Can you see that the point point of intersection where both x and y are positive is the one that's smaller than 4?

Hence you need to pick the root that's 4 - something (smaller than 4) and not 4 + something (bigger than 4).
16. (Original post by Zacken)
Your sketch is like this:

Can you see that the point point of intersection where both x and y are positive is the one that's smaller than 4?

Hence you need to pick the root that's 4 - something (smaller than 4) and not 4 + something (bigger than 4).
Thank you! I really appreciate the help, I get it now.
17. (Original post by Zacken)
The bit I've bolded. Don't do this.

Move everything over to the one side instead and then factorise an out.
Oopsie, of course - that removes a solution of x = 0. My mistake! If you were to do that, (genuine question here, if you can answer it for me at all), would you still be okay as long as you said x=0 is a possible solution? (though in this case, it's not as x must be a positive value?)
18. (Original post by backtraced)
Oopsie, of course - that remove a solution of x = 0. My mistake! If you were to do that, (genuine question here, if you can answer it for me at all), would you still be okay as long as you said x=0 is a possible solution? (though in this case, it's not as x must be a positive value?)
Yep, if you say (x=0) is a solution, and then divide by x, that's okay.
19. (Original post by frostyy)
Thank you! Just to be clear on this though, I thought that if I have x outside the brackets on one side of the equation, I should divide everything on the other side by it to take it out?

So x(4-x) = x^2(7-x)
4 - x = (x^2 / x)((7 - x) / x)

I understand that I'm wrong now though.
The main issue is, is that if you expand (x^2 / x)((7-x) / x), you'd get x^2(7- x)/x^2. You have to multiply back both the numerator and denominator - so the expansion wouldn't give you what you originally had, if that makes sense. It would look like you're dividing by x^2, not x. Forgive me if i'm wrong with any of this!
20. (Original post by backtraced)
The main issue is, is that if you expand (x^2 / x)((7-x) / x), you'd get x^2(7- x)/x^2. You have to multiply back both the numerator and denominator - so the expansion wouldn't give you what you originally had, if that makes sense. It would look like you're dividing by x^2, not x. Forgive me if i'm wrong with any of this!
Thanks! I understand now. I appreciate the help.

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