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

In summary: I don't 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
  • #1
chwala
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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
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}##
 
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  • #2
chwala said:
herefore, the equation takes the form,
c3x2−b(b2−3ac)x+a3=0
Take a look at sign of term x.
 
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  • #3
anuttarasammyak said:
Take a look at sign of term x.
that is where i am not getting...why is it ##+## instead of ##-##
 
  • #4
[tex](x-\alpha^{-3})(x-\beta^{-3})=x^2-(\alpha^{-3}+\beta^{-3})x+\alpha^{-3}\beta^{-3}[/tex]
 
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  • #5
anuttarasammyak said:
[tex](x-\alpha^{-3})(x-\beta^{-3})=x^2-(\alpha^{-3}+\beta^{-3})x+\alpha^{-3}\beta^{-3}[/tex]
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?
 
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  • #6
chwala said:
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.
 
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  • #7
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.
 
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  • #8
PeroK said:
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.
 
  • #9
etotheipi said:
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
 
  • #10
$$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{(\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)}{(\alpha \beta)^3} = \frac{(\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta)}{(\alpha \beta)^3}$$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.
 
  • #11
etotheipi said:
$$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{(\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)}{(\alpha \beta)^3} = \frac{(\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta)}{(\alpha \beta)^3}$$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}##
 
  • #12
chwala said:
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.
 
  • #13
chwala said:
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.
 
  • #14
chwala said:
you have just re written my equation, that's what i got...check

Cool. So what are we still doing here? :wink:

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.
 
  • #15
PeroK said:
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
 
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  • #16
chwala said:
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##.
 
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  • #17
PeroK said:
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.. :biggrin:
 
  • #18
1593439608110.png
 
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  • #19
1593439661674.png
 
  • #20
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.
 
  • #21
Example 6 is also correct.
 
  • #22
I will need to relax and check this again...
 
  • #23
chwala said:
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.
 
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  • #24
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:cool:
 
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  • #25
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 $$f(x) = (x-\alpha_1)(x - \alpha_2)\dots (x-\alpha_n)$$
 
  • #26
so i was just right after all...take your vote:smile:
 
  • #27
chwala said:
##\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}##
 
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  • #28
haruspex said:
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.
 
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  • #29
1593582929694.png


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

1593583056745.png
 
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  • #30
The approach below is from a colleague, let me know of any other approach to this part of the question.

1593583162924.png
 
  • #31
chwala said:
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.
 
  • #32
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!
 
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1. What is a quadratic equation?

A quadratic equation is a polynomial equation of the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. It is a second-degree polynomial and can be solved using various methods, including factoring, completing the square, and using the quadratic formula.

2. What are the roots of a quadratic equation?

The roots of a quadratic equation are the values of x that make the equation equal to 0. In other words, they are the solutions to the equation. A quadratic equation can have two, one, or zero real roots, depending on the value of the discriminant (b^2 - 4ac).

3. How can a quadratic equation be solved by finding the sum and product of its roots?

A quadratic equation can be solved by finding the sum and product of its roots using the Vieta's formulas. These formulas state that for a quadratic equation of the form ax^2 + bx + c = 0, the sum of the roots is -b/a and the product of the roots is c/a. By substituting these values into the equation, we can solve for the roots.

4. Can all quadratic equations be solved by finding the sum and product of its roots?

Yes, all quadratic equations can be solved by finding the sum and product of its roots using the Vieta's formulas. However, this method may not always be the most efficient or practical, especially for equations with irrational or complex roots.

5. What are the advantages of solving a quadratic equation by finding the sum and product of its roots?

Solving a quadratic equation by finding the sum and product of its roots can be advantageous because it is a straightforward and systematic method. It also allows for the quick determination of the number of real roots and their approximate values. Additionally, it can be used to solve more complex equations, such as cubic and quartic equations, by using Vieta's formulas for those equations as well.

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