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).

 

[Gavran]

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.

This can also be shown in the following way.

$\begin{align} &\arctan\frac\alpha c+\arctan\alpha=\arctan c\nonumber\\ &-\arctan\frac{\alpha}{\alpha\beta}+\arctan\alpha=\arctan c\nonumber\\ &-\arctan\frac1\beta+\arctan\alpha=\arctan c\nonumber\\ &\arctan\alpha-(\frac\pi2-\arctan\beta)=\arctan c\nonumber\\ &\arctan\alpha+\arctan\beta=\frac\pi2+\arctan c\nonumber\\ &\arctan\frac{\alpha+\beta}{1-\alpha\beta}=\arctan(-\frac1c)\nonumber\\ &\frac{p}{1+c}=-\frac{1}{c}\nonumber\\ &pc+c+1=0\nonumber\\ \end{align}$

 

[neilparker62]

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

The roots of the equation in part (i) are the same as the roots in part (ii). Since in both cases we have equations of the form ax^2 + bx + c = 0 with a=1 , 'c' = -c we must also have -p=(1+c)/c.