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How does this equal this?

@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.

Reply 1

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)

Reply 2

Zacken

Original post by Someboady

Common denominator and then use

\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

Reply 3

Someboady

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 :/

Reply 4

Someboady

Original post by Zacken

Common denominator and then use

\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

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.

Reply 5

KaylaB

Original post by Someboady

There you go, feel free to ask if you don't understand any parts

Reply 6

Zacken

Original post by Someboady

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.

$2\sqrt{1-\frac{x^2}{4^2}} = 2\sqrt{\frac{4-x^2}{x^2}} = 2 \times \frac{\sqrt{4-x^2}}{4} = 2 \times \frac{\sqrt{4-x^2}}{\sqrt{4}} = 2\times \frac{\sqrt{4-x^2}}{2}$

(edited 7 years ago)

Reply 7

cheesedrew

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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Reply 8

Someboady

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! :smile:

Reply 9

KaylaB

Original post by Someboady

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! :smile:

Yeah exactly! No problem :hat2:

Reply 10

FatShit

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

Reply 11

IrrationalRoot

Original post by Someboady

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! :smile:

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.).

Reply 12

Someboady

Original post by IrrationalRoot

I see. I initially thought they multiplied through by the 4 in x^2 / 4. Would this be allowed within a square root function?

Reply 13

IrrationalRoot

Original post by Someboady

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.
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