- #1

chwala

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- Homework Statement
- See attached

- Relevant Equations
- sum/product

For part (i),

##(x-α)(x-β)=x^2-(α+β)x+αβ##

##α+β = p## and ##αβ=-c##

therefore,##α^3+β^3=(α+β)^3-3αβ(α+β)##

=##p^3+3cp##

=##p(p^2+3c)##

For part (ii),

We know that; ##tan^{-1} x+tan^{-1} y##=##tan^{-1}\left[\dfrac {tan^{-1} x+tan^{-1} y}{1-tan^{-1} x⋅tan^{-1} y}\right]## then it follows that,

##tan^{-1}\left[ \frac {x}{c}\right]+tan^{-1} x##=##tan^{-1}\left[\dfrac {\dfrac{x}{c} + x}{1- \frac{x}{c}⋅ x}\right]##

We now have;

##tan^{-1}\left[\dfrac {\dfrac{x}{c} + x}{1- \frac{x}{c}⋅ x}\right]##=## tan^{-1}c##

##\left[\dfrac {\dfrac{x}{c} + x}{1- \frac{x}{c}⋅ x}\right]##=##c##

##\left[\dfrac {\dfrac{x}{c} + x}{1- \frac{x^2}{c}}\right]##=##c##

##\dfrac {x}{c}##+##x##=##c####(1##-##\dfrac{x^2}{c})##...from this we get,

##x^2+(\dfrac {1}{c}+1)x-c=0##

We know that, ##α+β = p## and ##αβ=-c##

it follows that

##-(β+α)##=##\frac {1}{c}+1## and ##αβ=-c##

then using,

##-(β+α)##=##\dfrac {1}{c}+1##

##-(β+α)##=##\dfrac {1+c}{c}##

##-p##=##\dfrac {1+c}{c}##

##-pc=1+c##

##⇒pc+c+1=0## Bingo,

I would appreciate any feedback on my steps...as i do not have markscheme or rather the solutions. Cheers guys.

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