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?
If x is positive, why take the negative root?
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 16042016 22:42

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 16042016 22:46
(Original post by frostyy)
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 16042016 22:48
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?
Post rating:1 
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 16042016 22:58
(Original post by Zacken)
From the sketch. What does the sketch look like? Does it show that x is less than 4 in anyway? 
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 16042016 23:01
(Original post by frostyy)
If you mean the sketches from the previous part, x was actually equal to 0, 4 and 7. 
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 16042016 23:02
(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? 
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 16042016 23:04
(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?

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 16042016 23:05
(Original post by frostyy)
That probably is the explanation. Thanks! 
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 16042016 23:07
(Original post by frostyy)
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 16042016 23:12
(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?
Anyhow, even if you did cancel, you should end up with
So that you have .Last edited by Zacken; 16042016 at 23:14. 
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 16042016 23:15
(Original post by Zacken)
Have you posted the wrong picture?

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 16042016 23:16
(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 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(4x) = x^2(7x)
DIVIDE BY X
(4x) = x(7x)
EXPAND
4x=7xx^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.Post rating:1 
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 16042016 23:18
(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(4x) = x^2(7x)
DIVIDE BY X
(4x) = x(7x)
EXPAND
4x=7xx^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. 
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 16042016 23:21
(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(4x) = x^2(7x)
DIVIDE BY X
(4x) = x(7x)
EXPAND
4x=7xx^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.
So x(4x) = x^2(7x)
4  x = (x^2 / x)((7  x) / x)
I understand that I'm wrong now though. 
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 16042016 23:21
(Original post by frostyy)
Forgive me, it's a bit late ;p.
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). 
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 16042016 23:22
(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). 
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 16042016 23:22
(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.Last edited by backtraced; 16042016 at 23:25. 
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 16042016 23:23
(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?)Post rating:1 
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 16042016 23:25
(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(4x) = x^2(7x)
4  x = (x^2 / x)((7  x) / x)
I understand that I'm wrong now though.Last edited by backtraced; 16042016 at 23:27. 
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 16042016 23:28
(Original post by backtraced)
The main issue is, is that if you expand (x^2 / x)((7x) / 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!Post rating:1
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Updated: April 16, 2016
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