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# Coordinate Systems

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1. Question:

Workings:

That is what I have ended up with. Why is this incorrect?
2. (Original post by NinCheng)
Question:

Workings:

That is what I have ended up with. Why is this incorrect?

You cannot do that with square roots, in general.

Instead, try factorising what you have under the square root sign.
3. (Original post by NinCheng)
Question:

Workings:

That is what I have ended up with. Why is this incorrect?
As above. Do you think that ? Surely not.

4. (Original post by 16Characters....)
You cannot do that with square roots, in general.

Instead, try factorising what you have under the square root sign.
Which leaves me with but now I'm clueless
5. (Original post by NinCheng)
Which leaves me with but now I'm clueless
Can you see how to factorise that bracket? You might like to let y = p^2
6. line 2, take a common factor of 9 out of expression which becomes 3 outside sqrt. The function left inside the sqrt is a perfect square.

Your fundamental error is that does NOT equal which is what your "solution' assumes.

if
did equal then try it with a and b = 1 which would predict that the square root of 2 was 2
7. (Original post by 16Characters....)
Can you see how to factorise that bracket? You might like to let y = p^2
Oh, so

8. (Original post by NinCheng)
Oh, so

Yep.
9. (Original post by 16Characters....)
Yep.
Thank you

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