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

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## 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:

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}##

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anuttarasammyak
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herefore, the equation takes the form,
c3x2−b(b2−3ac)x+a3=0
Take a look at sign of term x.

Mr Chanda
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Take a look at sign of term x.
that is where i am not getting...why is it ##+## instead of ##-##

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

PeroK
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$$(x-\alpha^{-3})(x-\beta^{-3})=x^2-(\alpha^{-3}+\beta^{-3})x+\alpha^{-3}\beta^{-3}$$
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 lets agree on one fact we are told that, ##∝## and ##β## are roots of the quadratic therefore their sum gives ##\frac {-b}{a}## right?

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PeroK
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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 lets 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.

Delta2
etotheipi
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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.

Delta2
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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.

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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
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$$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{(\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)}{(\alpha \beta)^3} = \frac{(\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta)}{(\alpha \beta)^3}$$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.

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$$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{(\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)}{(\alpha \beta)^3} = \frac{(\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta)}{(\alpha \beta)^3}$$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, thats 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}##

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you have just re written my equation, thats 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
Homework Helper
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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
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you have just re written my equation, thats 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.

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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
PeroK
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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##.

etotheipi
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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..

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Delta2
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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
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Example 6 is also correct.

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

PeroK
Homework Helper
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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.

etotheipi
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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 thats where the negatives cancel.
bingo

etotheipi and PeroK
etotheipi
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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 $$f(x) = (x-\alpha_1)(x - \alpha_2)\dots (x-\alpha_n)$$