The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

If x is positive, why take the negative root?

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?

Scroll to see replies

Reply 1
Avatar for Zacken
Zacken
22
Original post by frostyy
...


From the sketch. What does the sketch look like? Does it show that x is less than 4 in anyway?
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?
Reply 3
Avatar for frostyy
frostyy
OP
17
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.
Reply 4
Avatar for Zacken
Zacken
22
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?
Reply 5
Avatar for frostyy
frostyy
OP
17
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!
Reply 6
Avatar for frostyy
frostyy
OP
17
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?
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!
Reply 8
Avatar for Zacken
Zacken
22
Original post by frostyy
...


Have you posted the wrong picture?
Reply 9
Avatar for Zacken
Zacken
22
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 x=0x =0.

Anyhow, even if you did cancel, you should end up with x(4−x)=x2(7−x)⇒4−x=x2x(7−x)=x(7−x)x(4-x) = x^2(7-x) \Rightarrow 4-x = \frac{x^2}{x}(7-x) = x(7-x)

So that you have 4−x=x(7−x)4-x = x(7-x).
(edited 8 years ago)
Reply 10
Avatar for frostyy
frostyy
OP
17
Original post by Zacken
Have you posted the wrong picture?


Forgive me, it's a bit late ;p.

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.
Reply 12
Avatar for Zacken
Zacken
22
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 xx out.
Reply 13
Avatar for Zacken
Zacken
22
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).
Reply 14
Avatar for frostyy
frostyy
OP
17
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.
Reply 15
Avatar for frostyy
frostyy
OP
17
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.
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 xx 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?)
(edited 8 years ago)
Reply 17
Avatar for Zacken
Zacken
22
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.
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!
(edited 8 years ago)
Reply 19
Avatar for frostyy
frostyy
OP
17
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.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you