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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
Attached Images IMG_20141218_164606950.jpg (94.2 KB, 8 views) December 18th, 2014, 05:37 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 732 Math Focus: Physics, mathematical modelling, numerical and computational solutions 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. December 18th, 2014, 06:00 PM   #3
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Quote:
 Originally Posted by Benit13 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 December 19th, 2014, 01:21 AM #4 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 732 Math Focus: Physics, mathematical modelling, numerical and computational solutions No problem  Tags addition, geometric, progression Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jiasyuen Algebra 1 May 1st, 2014 03:40 AM jiasyuen Algebra 7 March 23rd, 2014 08:29 PM jiasyuen Algebra 1 November 19th, 2013 02:13 AM naman Algebra 1 March 5th, 2013 06:26 AM MIDI Elementary Math 10 January 23rd, 2011 05:09 AM

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