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December 18th, 2014, 02:25 AM   #1
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addition of geometric progression

how the addition of geometric series gives 2 and product gives 40
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December 18th, 2014, 05:37 AM   #2
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Short answer: The 2 and the 40 are determined from the coefficient of the $\displaystyle x^3$ term and the constant term in the quartic.

Long answer:

If the roots of the quartic are

$\displaystyle a-3d, a-d, a+d, a+3d$

then by multiplying (x-root) brackets together we should get the original quartic:

i.e. $\displaystyle (x - (a-3d))(x - (a-d))(x - (a+d))(x - (a+3d)) = x^4 -2x^3 - 21x^2 + 22x^3 + 40$

If we multiply out this whole left-hand side, we can collect up all of the $\displaystyle x^3$-terms, $\displaystyle x^2$-terms and $\displaystyle x$-terms to equate the coefficients with $\displaystyle a$ and $\displaystyle d$.

I'm not going to multiply out the whole thing, but I know that part of it is

$\displaystyle (x - (a-3d))(x - (a-d))(x - (a+d))(x - (a+3d)) = x^4 + (-(a-3d)-(a-d)-(a+d)-(a+3d))x^3 + ... + (-(a-3d))(-(a-d))(-(a+d))(-(a+3d))$

The coefficient in front of the $\displaystyle x^3$ term is

$\displaystyle (-(a-3d)-(a-d)-(a+d)-(a+3d)) = -4a$

If we compare this with the actual one, which is -2, we know that $\displaystyle a$ must be 1/2. Note that as a consequence of multiplying out the brackets we have ended up with a coefficient that is calculated by summing up all of the roots and placing a minus sign in front.

Also, the last term, the constant, is
$\displaystyle (-(a-3d))(-(a-d))(-(a+d))(-(a+3d))$
$\displaystyle = (a-3d)(a-d)(a+d)(a+3d)$
$\displaystyle = (a-3d)(a+3d) \cdot (a-d)(a+d)$
$\displaystyle = \left(a^2 - 9d^2\right)\left(a^2-d^2\right)$

Note that as a consequence of multiplying out the brackets we have ended up with a constant term that is the product of our roots. Comparing this with the actual constant term, which is 40, gives

$\displaystyle \left(a^2 - 9d^2\right)\left(a^2-d^2\right) = 40$.

We can also substitute the a=1/2 that we know, so

$\displaystyle \left(\left(\frac{1}{2}\right)^2 - 9d^2\right)\left(\left(\frac{1}{2}\right)^2-d^2\right) = 40$
$\displaystyle \left(\frac{1}{4} - 9d^2\right)\left(\frac{1}{4}-d^2\right) = 40$
$\displaystyle \left(1 - 36d^2\right)\left(1-4d^2\right) = 640$
$\displaystyle 1 - 40d^2 + 144d^4 = 640$
$\displaystyle 144d^4 - 40d^2 - 639 = 0$

Solving this gives $\displaystyle d = \pm 3/2$, as explained in the textbook. Now we know $\displaystyle a$ and $\displaystyle d$, we know all the solutions.
Thanks from Bhuvaneshnick

Last edited by Benit13; December 18th, 2014 at 05:40 AM.
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December 18th, 2014, 06:00 PM   #3
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Quote:
Originally Posted by Benit13 View Post
Short answer: The 2 and the 40 are determined from the coefficient of the $\displaystyle x^3$ term and the constant term in the quartic.

Long answer:

If the roots of the quartic are

$\displaystyle a-3d, a-d, a+d, a+3d$

then by multiplying (x-root) brackets together we should get the original quartic:

i.e. $\displaystyle (x - (a-3d))(x - (a-d))(x - (a+d))(x - (a+3d)) = x^4 -2x^3 - 21x^2 + 22x^3 + 40$

If we multiply out this whole left-hand side, we can collect up all of the $\displaystyle x^3$-terms, $\displaystyle x^2$-terms and $\displaystyle x$-terms to equate the coefficients with $\displaystyle a$ and $\displaystyle d$.

I'm not going to multiply out the whole thing, but I know that part of it is

$\displaystyle (x - (a-3d))(x - (a-d))(x - (a+d))(x - (a+3d)) = x^4 + (-(a-3d)-(a-d)-(a+d)-(a+3d))x^3 + ... + (-(a-3d))(-(a-d))(-(a+d))(-(a+3d))$

The coefficient in front of the $\displaystyle x^3$ term is

$\displaystyle (-(a-3d)-(a-d)-(a+d)-(a+3d)) = -4a$

If we compare this with the actual one, which is -2, we know that $\displaystyle a$ must be 1/2. Note that as a consequence of multiplying out the brackets we have ended up with a coefficient that is calculated by summing up all of the roots and placing a minus sign in front.

Also, the last term, the constant, is
$\displaystyle (-(a-3d))(-(a-d))(-(a+d))(-(a+3d))$
$\displaystyle = (a-3d)(a-d)(a+d)(a+3d)$
$\displaystyle = (a-3d)(a+3d) \cdot (a-d)(a+d)$
$\displaystyle = \left(a^2 - 9d^2\right)\left(a^2-d^2\right)$

Note that as a consequence of multiplying out the brackets we have ended up with a constant term that is the product of our roots. Comparing this with the actual constant term, which is 40, gives

$\displaystyle \left(a^2 - 9d^2\right)\left(a^2-d^2\right) = 40$.

We can also substitute the a=1/2 that we know, so

$\displaystyle \left(\left(\frac{1}{2}\right)^2 - 9d^2\right)\left(\left(\frac{1}{2}\right)^2-d^2\right) = 40$
$\displaystyle \left(\frac{1}{4} - 9d^2\right)\left(\frac{1}{4}-d^2\right) = 40$
$\displaystyle \left(1 - 36d^2\right)\left(1-4d^2\right) = 640$
$\displaystyle 1 - 40d^2 + 144d^4 = 640$
$\displaystyle 144d^4 - 40d^2 - 639 = 0$

Solving this gives $\displaystyle d = \pm 3/2$, as explained in the textbook. Now we know $\displaystyle a$ and $\displaystyle d$, we know all the solutions.

Thank you now i understood
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December 19th, 2014, 01:21 AM   #4
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No problem
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