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Think about it. If I put the negative root as a possibility, once I square both sides, that negative is gone leaving you with the same result as the positive possibility.

ax^3+cx+d=0\\ \therefore 3ax^2+c=0\ (Will\ share\ a\ root)\\ x^2=\frac{-c}{3a}\\\\ x=\sqrt{\frac{-c}{3a}} We can then substitute this root into the original equation. a(\sqrt{\frac{-c}{3a}})^3+c(\sqrt{\frac{-c}{3a}})=-d\\ Factorising.. \sqrt{\frac{-c}{3a}}(\frac{-c}{3}+c)=-d\\ Squaring\ both\...

Maybe the answers are wrong. I can't seem to get this either. And the formula is correct, so maybe your variables could also be wrong?

Cambridge for sure.

That method is fine, except you'd have to put a statement at the end stating that as LHS=RHS when x=0 and x=1, then the equations meet at these x values. Your method is basically a different way of doing the same thing. HOWEVER! - Considering that you've eliminated the 'y' variable from the...

Complex Numbers Polynomials Conics And 2 of the harder 3 unit topics.

Sorry, just managed to get LaTeX working: Part A: LHS=\frac{sinA+2sinAcosA}{1+cosA+cos^2A-sin^2A}\\\\%20=\frac{sinA+2sinAcosA}{1+cosA+cos^2A-1+cos^2A}\\\\%20=\frac{sinA(2cosA+1)}{cosA(2cosA+1)}\\\\%20=tanA\\\\%20=RHS Part B...

Hahah good work. If all else fails: (5^1/3+2)^-1 All over 1. Maybe it'll get you a sympathy mark or something :)

Ok part 2. LHS=(sin2x)(1-cosx)/(cosx)(1-cos2x) =(2sinxcosx)(1-cosx)/(cosx)(1-cos^2x+sin^2x) =(2sinx)(1-cosx)/(1-cos^2x+sin^2x) =(2sinx)(1-cosx)/(2sin^2x) =(1-cosx)/(sinx) So far I've just used sin2x, cos2x, 1-cos^2x substitutions to simplify. Now substituting cosx=(1-t^2)/(1+t^2), and...

Are those two separate questions? I hope so..haha. Here's the first: LHS= (SinA+2SinACosA)/(1+cosA+cos^2A-sin^A) =(SinA+2SinAcosA)/(1+cosA+cos^A-1+cos^2A) =SinA/CosA * (2CosA+1)/(2CosA+1) (Factorising the last line) =TanA =RHS Brb and ill do part two. Providing thats a separate...

Showing that those equations have identical y values for particular x values (ie subbing into the equations separately) and visa versa is fine and accepted. The other way would be to equate your two equations...goodluck with that haha :) So yeah, that would be fine.

Yes that would be fine. In that kind of question, they're just looking to see that you recognise the structure of the parametrics. You could state that the parametrics are x=acost, y=bsint. And therefore a=...etc But it would make no difference in terms of marks.

Ok this method doesn't use the proper formulas, but personally, its easier just to understand what's happening rather than just using a formula. Let a=First term of your arithmetic sequence. Let b= The common difference between them. a+(a+b)+(a+2b)=24 3a+3b=24 a+b=8 --- Equation 1...

I can relate. For a nerd at a selective school, whilst school is obviously the main focus, the arg error also remains a particular sideline goal.