<s>[INST] Solving the equation $11x^2-6000x-27500 =0$, preferably without the quadratic formula I obtained this form while solving an aptitude question.
$$\frac{3000}{x-50} + \frac{3000}{x+50} = 11$$
I changed it into quadratic equation
$$11x^2 -6000x - 27500 =0$$
but I don't know how to solve it.
I can't find two factor for 303500 that sums to 6000 or when I use formula the numbers become huge...
Without using calculator
how to solve it? is there any other simple way to solve [other method]? [or finding factor] I'm a beginner in math. Please explain your answer for me.
 [/INST] $$11x^2 - 6000x - 27500 = 0$$
$a=11$
$b=-6000 = 2^4 \cdot 3 \cdot 5^3$
$c = -27500 = 2^2 \cdot 5^4 \cdot 11$
$ac=-2^2 \cdot 5^4 \cdot 11^2$
Since $ac$ is negative, we will search for integers $u$ and $v$ such that their product is $=2^2 \cdot 5^4 \cdot 11^2$ and their difference is $2^4 \cdot 3 \cdot 5^3 = 6000$. We will sort out exactly where the plusses and minusses go later.
We will try to guess at these numbers by looking at the factors and using a little logic.
If $11 \mid u$ and $11 \mid v$, then $11 \mid u+v$. So both $11's$ must belong to $u$ or $v$ but not both. So we start with this.
\begin{array}{l|c|c|}
        & u & v \\
\hline
\text{powers of 2}  &       &  \\
\text{powers of 5}  &       &  \\
\text{powers of 11} &  11^2 & 1\\
\hline
\text{product} & 121\\
\hline
\end{array}
Since $2 \mid u+v$ and $2 \mid uv$, then $u$ and $v$ must each have at least one factor of $2$. So we get this.
\begin{array}{l|c|c|}
        & u & v \\
\hline
\text{powers of 2}  &   2   & 2\\
\text{powers of 5}  &       &  \\
\text{powers of 11} &  11^2 & 1\\
\hline
\text{product} & 242 & 2\\
\hline
\end{array}
Similarly, $u$ and $v$ must share at least one $5$. So where do we put the other two? Since $6000$ is a pretty big number, we will try putting three of the $5's$ on the same side as the $11's$.
\begin{array}{l|c|c|}
        & u & v \\
\hline
\text{powers of 2}  &   2   & 2 \\
\text{powers of 5}  &   5^3 & 5 \\
\text{powers of 11} &  11^2 & 1 \\
\hline
\text{product} & 30250 & 10\\
\hline
\end{array}
And $u-v$ isn't $6000$.
So let's move one of the $5's$ over.
\begin{array}{l|c|c|}
        & u & v \\
\hline
\text{powers of 2}  &   2   & 2 \\
\text{powers of 5}  &   5^2 & 5^2 \\
\text{powers of 11} &  11^2 & 1 \\
\hline
\text{product} & 6050 & 50\\
\hline
\end{array}
And $u-v=6000$.
Using the $ac$ method, we get
$$\dfrac{(11x-6050)}{11} \dfrac{(11x+50)}{1}$$
$$(x-550)(11x+50)$$
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