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?
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, $\displaystyle -\dfrac{\dfrac{295}{3}}{\dfrac{1}{6}} = -6\dfrac{295}{3} = -590$
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.
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.