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# How does this equal this?

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1. Zacken

Sorry I don't quite follow. This is from a June 2013 C3 (R) Paper Q 5 part C.
Arsey's mark scheme got to the LHS of the equation and the Mark scheme gave the RHS of this equation so i'm wondering hwo you get from Arsey's to the Mark schemes.
2. first of all on the left, square the (x/2) out to get (X^2)/4now times everything in the root by 4 to get 4-X^2(this eliminates the 2 from outside the root because 4^0.5 i.e square root 4 is two)

Zacken

Sorry I don't quite follow. This is from a June 2013 C3 (R) Paper Q 5 part C.
Arsey's mark scheme got to the LHS of the equation and the Mark scheme gave the RHS of this equation so i'm wondering hwo you get from Arsey's to the Mark schemes.
Common denominator and then use
4. (Original post by cheesedrew)
first of all on the left, square the (x/2) out to get (X^2)/4now times everything in the root by 4 to get 4-X^2(this eliminates the 2 from outside the root because 4^0.5 i.e square root 4 is two)
I don't quite follow the last step. Why exactly does it eliminate the 2 outside the square root :/
5. (Original post by Zacken)
Common denominator and then use

Sorry how do I go from L.H.S to R.H.S without using the R.H.S of the equation.

I don't understand what happens to the 2 at the bottom.

Zacken

Sorry I don't quite follow. This is from a June 2013 C3 (R) Paper Q 5 part C.
Arsey's mark scheme got to the LHS of the equation and the Mark scheme gave the RHS of this equation so i'm wondering hwo you get from Arsey's to the Mark schemes.

There you go, feel free to ask if you don't understand any parts
Sorry how do I go from L.H.S to R.H.S without using the R.H.S of the equation.

I don't understand what happens to the 2 at the bottom.
I don't quite see the problem.

$2\sqrt{1-\frac{x^2}{4^2}} = 2\sqrt{\frac{4-x^2}{x^2}} = 2 \times \frac{\sqrt{4-x^2}}{4} = 2 \times \frac{\sqrt{4-x^2}}{\sqrt{4}} = 2\times \frac{\sqrt{4-x^2}}{2}$
8. it gets red of the 2 at the bottom because of this rule: a^0.5 multiplied by b^0.5 is equal to (a*b)^0.5
e.g root of 2 multiplied by root of 3 is the same at the root of 6

when you're multiplying out the bottom, what you're actually doing is multiplying by root of 4
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9. (Original post by KaylaB)

There you go, feel free to ask if you don't understand any parts
Ah I get it, so you rewrite 2 as sqrt 4 and then you can use rules of surds to multiply through. Got it! Thanks a bunch dudette!
Ah I get it, so you rewrite 2 as sqrt 4 and then you can use rules of surds to multiply through. Got it! Thanks a bunch dudette!
Yeah exactly! No problem
11. First I dont do a-level maths, i am currently doing gcse. So i may be wrong. But i would assume square the (x/2) out to get (X^2)/4now times everything in the root by 4 to get 4-X^2 therefore this eliminates the 2 from outside the root because 4^0.5
Ah I get it, so you rewrite 2 as sqrt 4 and then you can use rules of surds to multiply through. Got it! Thanks a bunch dudette!
Yes this is the best way to do it. It's a very useful and common thing to know that you can 'put an outside number in the square root' by putting it's square in the square root (and correspondingly for cube roots etc.).
13. (Original post by IrrationalRoot)
Yes this is the best way to do it. It's a very useful and common thing to know that you can 'put an outside number in the square root' by putting it's square in the square root (and correspondingly for cube roots etc.).
I see. I initially thought they multiplied through by the 4 in x^2 / 4. Would this be allowed within a square root function?
I see. I initially thought they multiplied through by the 4 in x^2 / 4. Would this be allowed within a square root function?
Yes because you write 2 as sqrt(4) and then you just have a square root with 4 multiplied by what was originally in the square root.
Latex is not working today...

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