The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

C1 surds

I was doing the jan 2013 paper and it seems like there is only one answer to this question... Isn't there supposed to be two.

5+5(root2) - 2(root2) -root16

Basically, we gotta express dis in the form a+b(root2), where a and b are integers.

So we have 5 + 3(root2) -root16

Now, if we take the root of 16 as -4, we get the answer to be 9 +3root2, but if we take the root of 16 as 4, we get the answer to be 1+3root2, which is the only answer in the mark scheme

Will you get the marks if you got 9 instead of 5 :hmmm:

For c1, when should we take both roots, and when should we just take the positive root :hmmm:
Posted from TSR Mobile
(edited 8 years ago)
Original post by life's a pain
x


See here.

The square root of something is just the positive number - it's when you stick a minus sign in front of it that you're 'taking the negative root'. So always take the positive root in situations like that.
Original post by SeanFM
See here.

The square root of something is just the positive number - it's when you stick a minus sign in front of it that you're 'taking the negative root'. So always take the positive root in situations like that.


:hmmm: i cant see the latex on ma phone. The latex just appears as smiley faces (:smile:)

I'll check it out when i go on ma laptop

But isnt the root of 16 plus or minus 4... I dont understand :hmmm:
Posted from TSR Mobile
Original post by life's a pain
:hmmm: i cant see the latex on ma phone. The latex just appears as smiley faces (:smile:)

I'll check it out when i go on ma laptop

But isnt the root of 16 plus or minus 4... I dont understand :hmmm:
Posted from TSR Mobile


The root of 16 is 4. Only one answer is given by the square root function, and that's the positive answer - never the negative.
Original post by SeanFM
The root of 16 is 4. Only one answer is given by the square root function, and that's the positive answer - never the negative.


Yes, looking at the graph of y = rootx, it is a one-to-one function

Nevver do we got two values of y for 1 value of x

So why have my stupid teachers always been saying there are two solutions :hmmm:

Posted from TSR Mobile
Original post by SeanFM
The root of 16 is 4. Only one answer is given by the square root function, and that's the positive answer - never the negative.


What about if we have

(x-4)^2 = 4

Then we got:

X-4 = +/- 2, no?

Posted from TSR Mobile
5+5(root2) - 2(root2) -root16
root16=4
: 5+4=1
:1+5root2-2root2

= 1+3root2
Original post by jambojim97
5+5(root2) - 2(root2) -root16
root16=4
: 5+4=1
:1+5root2-2root2

= 1+3root2


Naaah

Mans asking why the root of 16 is taken as 4 rather than +/- 4

Posted from TSR Mobile
Original post by life's a pain
What about if we have

(x-4)^2 = 4

Then we got:

X-4 = +/- 2, no?

Posted from TSR Mobile


Yes, there can be 2 solutions.

In the example you've given that is what is practied, but it looks like there's an intermediate step to get around 4^2 not giving -2.

This could be horribly misleading, but you could stick a modulus sign around (x-4) when you square root both sides (where the modulus of a number, p say, is written as l p l , which is the highest value of +p and -p (so always returns a positive number)) .

So you have l x - 4 l = 2, so either x - 4 = 2, or 4 - x = 2 (where x - 4 is p and 4 - x is -p), and when it's 4 - x you could write that as -(x-4) = 2, which is (x-4) = -2, and of course the other solution comes from x - 4 = 2.
I think my teacher said something along the lines of:
If you are square rooting an unknown then it has to be +/-
If you are square rooting a constant then it will be the positive root
Original post by SeanFM
Yes, there can be 2 solutions.

In the example you've given that is what is practied, but it looks like there's an intermediate step to get around 4^2 not giving -2.

This could be horribly misleading, but you could stick a modulus sign around (x-4) when you square root both sides (where the modulus of a number, p say, is written as l p l , which is the highest value of +p and -p (so always returns a positive number)) .

So you have l x - 4 l = 2, so either x - 4 = 2, or 4 - x = 2 (where x - 4 is p and 4 - x is -p), and when it's 4 - x you could write that as -(x-4) = 2, which is (x-4) = -2, and of course the other solution comes from x - 4 = 2.


Aaaaah, i see

So basically when i go from

(X-4)^2 = 4 to x-4 = +/-2, we've got some next methods and working out to get the -2, but this working out is always omitted

Why is this working out never shown, and how exactly do you get -2?

Posted from TSR Mobile
Original post by BasicMistake
I think my teacher said something along the lines of:
If you are square rooting an unknown then it has to be +/-
If you are square rooting a constant then it will be the positive root


Thank you my friend

But i really want to understand why, and not just remember rules and algorithms :hmmm:

Cos maybe then i will forget in the exam :yes:
Posted from TSR Mobile
Original post by life's a pain
Aaaaah, i see

So basically when i go from

(X-4)^2 = 4 to x-4 = +/-2, we've got some next methods and working out to get the -2, but this working out is always omitted

Why is this working out never shown, and how exactly do you get -2?

Posted from TSR Mobile


It could be something that I've made up - in which case I would delete the posts, but it seems like a way of getting to the -2.

I don't remember being taught such a thing at A-level so I suppose the +/- 2 bit is just taken for granted.

I've shown you a way of getting to -2 already, I guess you'll understand it better when you're introduced to the modulus function (at some point in A2 perhaps).
Original post by SeanFM
It could be something that I've made up - in which case I would delete the posts, but it seems like a way of getting to the -2.

I don't remember being taught such a thing at A-level so I suppose the +/- 2 bit is just taken for granted.

I've shown you a way of getting to -2 already, I guess you'll understand it better when you're introduced to the modulus function (at some point in A2 perhaps).


Yeh i did my a2s already :yes:

And i am just retaking c1 now

So i understand the modulus stuff

But why would you randomly introduce modulus.. Is this acceptable :hmmm:

Posted from TSR Mobile
Original post by life's a pain
Thank you my friend

But i really want to understand why, and not just remember rules and algorithms :hmmm:

Cos maybe then i will forget in the exam :yes:
Posted from TSR Mobile

It's a matter of terminology.

http://www.thestudentroom.co.uk/showthread.php?t=2912377&page=2&p=50879387#post50879387

EDIT: remember, the square root function *is a function*. That is, given any input (eg. 4) it has a unique output (2). When you use it to solve equations, though, you lose solutions: you can't go from (x^2 = 4) x2=4x^2 = 4 to (x = sqrt 4 = 2) x=4=2x=\sqrt{4} = 2, because you've lost a solution. The solution you've lost is the solution (x=- sqrt(4)) x=−4x=-\sqrt{4}. In the same way as you can't go from (sin x = 1) sin⁡(x)=1\sin(x) = 1 to (x = pi/2) x=π2x=\frac{\pi}{2} because you've lost solutions: this time, you've lost many more solutions, of the form (pi/2 + 2 pi n) π2+2πn\frac{\pi}{2} + 2 \pi n, for instance.

EDIT EDIT: added plaintext rather than LaTeX above
(edited 8 years ago)
Original post by Smaug123
It's a matter of terminology.

http://www.thestudentroom.co.uk/showthread.php?t=2912377&page=2&p=50879387#post50879387

EDIT: remember, the square root function *is a function*. That is, given any input (eg. 4) it has a unique output (2). When you use it to solve equations, though, you lose solutions: you can't go from x2=4x^2 = 4 to x=4=2x=\sqrt{4} = 2, because you've lost a solution. The solution you've lost is the solution x=−4x=-\sqrt{4}. In the same way as you can't go from sin⁡(x)=1\sin(x) = 1 to x=π2x=\frac{\pi}{2} because you've lost solutions: this time, you've lost many more solutions, of the form π2+2πn\frac{\pi}{2} + 2 \pi n, for instance.


I defo need to go on ma laptop soon, as i cant see the latex

Thanks mate, hopefully then i'll be able to see and understand the latex

Posted from TSR Mobile
Original post by life's a pain
I defo need to go on ma laptop soon, as i cant see the latex

Thanks mate, hopefully then i'll be able to see and understand the latex

Posted from TSR Mobile

Edited plain text in.
I see. I understand now

thanks to y'alls. i appreciate it :shakehand:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you