Solving a quadratic equation as a sum and product of its roots

[chwala]

Homework Statement

If $∝$ and $β$ are the roots of the equation $ax^2+bx+c=0$, obtain the equation whose roots are $\frac {1}{∝^3}$ and $\frac {1}{β^3}$

Relevant Equations

quadratic equations

for the sum,
$\frac {1}{∝^3}$+$\frac {1}{β^3}$=$\frac {β^3+∝^3}{∝^3β^3}$
=$\frac {(∝+β)[(∝+β)^2-3∝β]}{∝^3β^3}$
=$\frac {-b}{a}$. $[\frac{b^2-3ac}{a^2}]$. $\frac {a^3}{c^3}$
for the product,

$\frac {1}{∝^3}$ . $\frac {1}{β^3}$=$\frac {a^3}{c^3}$
therefore, on multiplying each term by $c^3$, the equation takes the form,
$c^3x^2-b(b^2-3ac)x+a^3=0$
now where my problem is that the textbook indicates that the answer is
$c^3x^2+b(b^2-3ac)x+a^3=0$ this is where i need clarification. In my understanding, i.e in reference to sum of roots
where if $m$ and $n$ are roots of $ax^2+bx+c=0$, then it follows that $m+n$=$\frac {-b}{a}$

 

[anuttarasammyak]

herefore, the equation takes the form,
c3x2−b(b2−3ac)x+a3=0

Take a look at sign of term x.

 

[chwala]

Take a look at sign of term x.

that is where i am not getting...why is it $+$ instead of $-$

 

[anuttarasammyak]

$$(x-\alpha^{-3})(x-\beta^{-3})=x^2-(\alpha^{-3}+\beta^{-3})x+\alpha^{-3}\beta^{-3}$$

 

[chwala]

$$(x-\alpha^{-3})(x-\beta^{-3})=x^2-(\alpha^{-3}+\beta^{-3})x+\alpha^{-3}\beta^{-3}$$

am still not getting, if you equate your coefficient of $x$ to the expression $ax^2+bx+c$ for the sum you will be getting $∝^{-3}+b^{-3}=\frac {-b}{a}$ which is what i did in my working...wait i check something here...still not seeing anything wrong, ok let's agree on one fact we are told that, $∝$ and $β$ are roots of the quadratic therefore their sum gives $\frac {-b}{a}$ right?

 

[PeroK]

am still not getting, if you equate your coefficient of $x$ to the expression $ax^2+bx+c$ for the sum you will be getting $∝^{-3}+b^{-3}=\frac {-b}{a}$ which is what i did in my working...wait i check something here...still not seeing anything wrong, ok let's agree on one fact we are told that, $∝$ and $β$ are roots of the quadratic therefore their sum gives $\frac {-b}{a}$ right?

I think you have done everything right but you've missed a negative sign somewhere. The error must be somewhere in the details.

 

[etotheipi]

Let the new equation be $px^2 + qx + r = 0$. You have, if you work it through, $$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{3abc - b^3}{c^3} = -\frac{q}{p}$$ and $$\frac{1}{\alpha^3 \beta^3} = \frac{a^3}{c^3} = \frac{r}{p}$$If you substitute those in, get rid of the irrelevant $p$ terms and multiply through by $c^3$ you should get the book answer.

By the way, my favourite approach to these problems is to, here for instance, apply the substitution $u = \frac{1}{x^3}$. That will give you the right answer straight off the bat.

 

[chwala]

I think you have done everything right but you've missed a negative sign somewhere. The error must be somewhere in the details.

the only reason from my reading is that when the coefficient of $x^2$ is unity ...then the coefficient of $x$=$-$( the sum of the roots)...no problem with the product part of the equation, i guess this is where the negatives cancel.

 

[chwala]

Let the new equation be $px^2 + qx + r = 0$. You have, if you work it through, $$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{3abc - b^3}{c^3} = -\frac{q}{p}$$ and $$\frac{1}{\alpha^3 \beta^3} = \frac{a^3}{c^3} = \frac{r}{p}$$If you substitute those in, get rid of the irrelevant $p$ terms and multiply through by $c^3$ you should get the book answer.

By the way, my favourite approach to these problems is to, here for instance, apply the substitution $u = \frac{1}{x^3}$. That will give you the right answer straight off the bat.

i do not understand what you are indicating by changing the variables to $a, b,c...$and so on...not necessary i think. look at my working, let me know where i have made a mistake. thanks

 

[etotheipi]

$$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{(\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)}{(\alpha \beta)^3} = \frac{(\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta)}{(\alpha \beta)^3}$$From the original equations we have that $\alpha \beta = \frac{c}{a}$ and $\alpha + \beta = -\frac{b}{a}$. So we can simplify further to $$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{(-\frac{b}{a})(\frac{b^2}{a^2} - \frac{3c}{a})}{\frac{c^3}{a^3}} = \frac{3abc - b^3}{c^3}$$ Also we have that $$\frac{1}{\alpha^3} \frac{1}{\beta^3} = \frac{a^3}{c^3}$$That is then sufficient to obtain the coefficients of the new equation in terms of the old ones, see #7.

 

[chwala]

$$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{(\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)}{(\alpha \beta)^3} = \frac{(\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta)}{(\alpha \beta)^3}$$From the original equations we have that $\alpha \beta = \frac{c}{a}$ and $\alpha + \beta = -\frac{b}{a}$. So we can simplify further to $$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{(-\frac{b}{a})(\frac{b^2}{a^2} - \frac{3c}{a})}{\frac{c^3}{a^3}} = \frac{3abc - b^3}{c^3}$$ Also we have that $$\frac{1}{\alpha^3} \frac{1}{\beta^3} = \frac{a^3}{c^3}$$That is then sufficient to obtain the coefficients of the new equation in terms of the old ones, see #7.

you have just re written my equation, that's what i got...check

=$\frac {(∝+β)[(∝+β)^2-3∝β]}{∝^3β^3}$
=$\frac {-b}{a}$. $[\frac{b^2-3ac}{a^2}]$. $\frac {a^3}{c^3}$

 

[chwala]

you have just re written my equation, that's what i got...check

=$\frac {(∝+β)[(∝+β)^2-3∝β]}{∝^3β^3}$
=$\frac {-b}{a}$. $[\frac{b^2-3ac}{a^2}]$. $\frac {a^3}{c^3}$

if you read my query my concern was why textbook indicates $+b^3$ instead of $-b^3$...but going with the reasoning on post 8, i might know why...as indicated on my remarks on post 8.

 

[PeroK]

if you read my query my concern was why textbook indicates $+b^3$ instead of $-b^3$...but going with the reason on post 8, i might know why...as indicated on my remarks on post 8.

The book is correct. Try $a = c = 1, b = -2$. The roots are $\alpha, \beta \pm 1$, so $\frac 1 {\alpha^3}, \frac 1 {\beta^3} = \pm 1$. And we should get the same equation. This shows that it must be $+b$.

You need to check the details of your calculations.

 

[etotheipi]

you have just re written my equation, that's what i got...check

Cool. So what are we still doing here?

The new equation is $px^2 + qx + r = 0$. We know that $q = p(\frac{b^3 - 3ac}{c^3})$ and $r = \frac{pa^3}{c^3}$, from post #7. Plug those in and cancel the $p$'s.

 

[chwala]

The book is correct. Try $a = c = 1, b = -2$. The roots are $\alpha, \beta \pm 1$, so $\frac 1 {\alpha^3}, \frac 1 {\beta^3} = \pm 1$. And we should get the same equation. This shows that it must be $+b$.

You need to check the details of your calculations.

my calculations are correct, i think the reasoning is on post 8...let me post a similar example with its worked out (textbook solution) so that we get clarity on this

 

[PeroK]

my calculations are correct, i think the reasoning is on post 8...let me post a similar example with its worked out (textbook solution) so that we get clarity on this

You're outvoted 4:1 on that. The book plus all three of us helping you here get $+b$.

 

[chwala]

You're outvoted 4:1 on that. The book plus all three of us helping you here get $+b$.

i am not in dispute, i just wanted to know why is it positive and not negative...here check this..

 

[chwala]

 

[chwala]

 

[chwala]

look at example 6...i used a similar approach and they've given a reason on the change of sign as earlier indicated in post 8.

 

[etotheipi]

Example 6 is also correct.

 

[chwala]

I will need to relax and check this again...

 

[PeroK]

look at example 6...i used a similar approach and they've given a reason on the change of sign as earlier indicated in post 8.

It would be great if there were a method by which you could look at an equation or formula and check (by some general rule) which terms should be $+$ and which terms should be $-$. There is no such rule. Just because the answer to one problem looks like $a -b + c$ doesn't mean that the answer to all similar problems looks like $a - b + c$. Maths doesn't work like that.

In this case, if you do the calculations correctly you get $+b(b^2 - 3ac)$. You might have done a problem last week where the answer was $-p(p^2 - 3q)$. That means nothing. Different problem, different solution.

 

[chwala]

I think i now understand it very well. first to indicate that my working is correct. Consider the pesky sum part of the quadratic equation, i managed to express the coefficient of $x$ as a sum of the roots as indicated in post 12.
Now we need to re-write the quadratic equation in terms of the sum and product of the roots, therefore (check textbook equation 1.4) the coefficient of $x$ is $-(∝+β)$ and that's where the negatives cancel.
bingo

 

[etotheipi]

Also in the general case of a polynomial of order $n$, the coefficient of $x^k$ will be a sum consisting of $^nC_k$ terms each of which is a possible distinct product of $(n-k)$ roots. If $(n-k)$ is odd, then you will need to insert a negative sign when relating this sum to the coefficient in the polynomial. It's perhaps easiest to see by looking at the combinatorics of the factorised form $$f(x) = (x-\alpha_1)(x - \alpha_2)\dots (x-\alpha_n)$$

 

[chwala]

so i was just right after all...take your vote

 

[haruspex]

$\frac {1}{∝^3}$+$\frac {1}{β^3}$=$\frac {-b}{a}$. $[\frac{b^2-3ac}{a^2}]$. $\frac {a^3}{c^3}$

therefore, on multiplying each term by $c^3$, the equation takes the form,
$c^3x^2-b(b^2-3ac)x+a^3=0$

If the sum of the roots is $\frac {-b}{a}$. $[\frac{b^2-3ac}{a^2}]$. $\frac {a^3}{c^3}$ then the coefficient of x in the equation will be $\frac {+b}{a}$. $[\frac{b^2-3ac}{a^2}]$. $\frac {a^3}{c^3}$

 

[chwala]

If the sum of the roots is $\frac {-b}{a}$. $[\frac{b^2-3ac}{a^2}]$. $\frac {a^3}{c^3}$ then the coefficient of x in the equation will be $\frac {+b}{a}$. $[\frac{b^2-3ac}{a^2}]$. $\frac {a^3}{c^3}$

yeah, now its very clear to me...just small detail that i had missed thanks Haruspex.

 

[chwala]

This was the second part of the question and i have used this approach below;

 

[chwala]

The approach below is from a colleague, let me know of any other approach to this part of the question.

 

[PeroK]

The approach below is from a colleague, let me know of any other approach to this part of the question.

View attachment 265589

Your solution is a lot neater than that.

 

[chwala]

Thanks Perok, it took me a while to think on that...all the passion for this great subject...its a pleasure. I will be posting and sharing more questions...bingo!