x is a real number.

x+8+x+10+x+12+.........5x-2+5x=578

How many times bigger is the multiplication of x values that supplies this equation than the sum of the x values?

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- April 16th 2014, 05:21 AMkastamonusum of roots
x is a real number.

x+8+x+10+x+12+.........5x-2+5x=578

How many times bigger is the multiplication of x values that supplies this equation than the sum of the x values? - April 16th 2014, 09:41 AMromsekRe: sum of roots
sorry I can't understand this at all.

- April 16th 2014, 10:01 AMkastamonuRe: sum of roots
First we will find the sum of x values that supplies this equation. After that we will find the multiplication of x values that supplies this equation and we will find the difference. Answer is -590 but I couldn't get it.

- April 16th 2014, 10:14 AMSlipEternalRe: sum of roots
It looks like the equation can be rewritten as $\displaystyle \sum_{k=4}^{2x} (x+2k) = 578$.

Breaking it down:

$\displaystyle \begin{align*}\sum_{k=4}^{2x} (x+2k) & = \sum_{k=4}^{2x}x + \sum_{k=4}^{2x}2k \\ & = x\sum_{k=4}^{2x}1 + 2\sum_{k=4}^{2x}k \\ & = x(2x-3) + 2\left(\sum_{k=1}^{2x}k - \sum_{k=1}^3 k\right) \\ & = x(2x-3) + 2\left(\dfrac{2x(2x+1)}{2} - \dfrac{3(4)}{2}\right) \\ & = x(2x-3) + 2x(2x+1)-12 \\ & = 6x^2-x-12 = 578\end{align*}$

So, you get a quadratic equation, and find $x = \dfrac{1\pm 119}{12}$. So, the sum of $x$-values is $10 - \dfrac{59}{6}= \dfrac{1}{6}$ and the product of them is $10 \left(-\dfrac{59}{6}\right) = -\dfrac{295}{3}$.

So, $\displaystyle -\dfrac{\dfrac{295}{3}}{\dfrac{1}{6}} = -6\dfrac{295}{3} = -590$ - April 16th 2014, 11:10 AMkastamonuRe: sum of roots
many thanks.

- April 17th 2014, 07:48 AMebainesRe: sum of roots
I'm trying to understand this equation:

for the case of x = -59/6. X=59/6 makes the series:

$ \frac {-59} 6 + 8 +\frac {-59} 6 + 10 + \frac {-59} 6 + 12 + ....$

It never reaches 5 x -59/6 = -49.1666

I think the problem is that the formula

$\displaystyle \sum _{k=1} ^{2x} = \frac {2x(2x+1)} 2 $

works only for positive integer values of 2x. So if we find x= -59/6 it doesn't hold up. I maintain that there is only one solution to the series, and that is x=10. - April 17th 2014, 08:06 AMSlipEternalRe: sum of roots
- April 17th 2014, 09:40 AMebainesRe: sum of roots
The queston didn't ask about a quadratic - it asked about a sum of a series, and that quadratic equation only defines the sum for positive integers of x. This is like asking what values of x satisfy x+1=5 and approaching it be squaring both sides, rearranging and getting the quadratic x^2 +2x-24 = 0, and concluding that the two possible values for x are 4 and -6. Oh well, it does appear that your method matched what the person who devides the question was expecting.