Solve the quadratic equation involving sum and product

[chwala]

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;

$tan^{-1} x+tan^{-1} y$=$tan^{-1}\left[\dfrac {x+y}{1- x⋅ y}\right]$ 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.

 

[Office_Shredder]

You didn't do the last part of (1) where you have to find a quadratic equation?

 

[chwala]

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,
$(x-α^3)(x-β^3)$
$=x^2-(α^3+β^3)x+α^3β^3$
$=x^2-(p^3+3pc)x-c^3$

 

[chwala]

Homework Statement: See attached
Relevant Equations: sum/product

View attachment 296980

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.

for part i. i guess the quadratic equation was not indicated. I will go ahead and finish on that; The quadratic equation will be $x^2 -p(p^2+3c)x^2-c^3=0$

 

[chwala]

You didn't do the last part of (1) where you have to find a quadratic equation?

done.

 

[pasmith]

This question seems poorly designed, in that part (i) has absolutely nothing to do with part (ii).