Solve the quadratic equation involving sum and product

AI Thread Summary
The discussion focuses on solving a quadratic equation involving the sum and product of roots. It establishes that for roots α and β, the relationships α + β = p and αβ = -c lead to the expression for α^3 + β^3 as p(p^2 + 3c). The second part involves manipulating the arctangent identities to derive a quadratic equation in terms of x, resulting in x^2 + (1/c + 1)x - c = 0. The final step confirms that the quadratic equation can be expressed as (x - α^3)(x - β^3) = x^2 - (p^3 + 3pc)x - c^3. Feedback is requested on the steps taken to arrive at these conclusions.
chwala
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Homework Statement
See attached
Relevant Equations
sum/product
1644637889605.png


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,:cool:
I would appreciate any feedback on my steps...as i do not have markscheme or rather the solutions. Cheers guys.
 
Last edited:
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You didn't do the last part of (1) where you have to find a quadratic equation?
 
For, the last part of (1), we shall have the factors;
##x=α^3## and ##x=β^3##, thus our quadratic equation will be of the form,
##(x-α^3)(x-β^3)##
##=x^2-(α^3+β^3)x+α^3β^3##
##=x^2-(p^3+3pc)x-c^3##
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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