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.
 A: Hint
Do a substitution $x = 50y$. Then the equation becomes
$$
\frac {3000}{50(y-1)} + \frac {3000}{50(y+1)} = 11 \\
\frac {60}{y-1} + \frac {60}{y+1} = 11
$$
which should be a bit easier to solve... I guess...
A: Multiply by 11, and replace $y=11x$. Then you get
$$y^2-6000y-302500=0 \,.$$
Now complete the square:
$$y^2-6000y+3000^2=3000^2+302500$$
Last:
$$3000^2+302500=3000\times 3000+3025\times 100=600 \times 5 \times 6 \times 500+121\times25\times100$$
$$=2500 \times (3600+121)=2500 \times 3721=50^2 \times 61^2$$
Thus you get
$$(y+3000)^2=3050^2$$
A: use this formula $ax^2+bx+c=0\implies x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
$$11x^2-6000x-27500=0$$
here $a=11,b=-6000,c=-27500$
just put these valuse in above formula and you got answer.
second approach:
$$11x^2-6000x-27500=0$$
$$11x^2-6050x+50x-27500=0$$
$$11x(x-550)+50(x-550)=0$$
$$(x-550)(11x+50)=0$$
$$(x-550)=0\;\;,(11x+50)=0$$
$$x=550,-\dfrac{50}{11}$$
A: There isn't any standard, guaranteed method apart from the quadratic formula to solve a quadractic equation. However sometimes there are "ad-hoc tricks" which might help you get one root.
The RHS of the equation is an integer; You might suspect that an $x$ such that both the terms on the LHS are integers might be a root (this does not have to be true at all, but it's not bad to try).
Also since $x-50$ and $x+50$ differ by $100$, you want a number $y$ such that both $y$ and $y+100$ divide $3000$. Noticing that $500$ and $600$ satisfy this gives $x=550$ as a root.
Using this, you can find the other root quite easily to be $x=-\frac{50}{11}$ since the product of the roots is $-27500/11$.
A: $$11x^2 - 6000x - 27500 = 0$$
$a=11$
$b=-6000 = 2^4 \cdot 3 \cdot 5^3$
$c = -27500 = 2^2 \cdot 5^4 \cdot 11$
$ac=-2^2 \cdot 5^4 \cdot 11^2$
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
$$\dfrac{(11x-6050)}{11} \dfrac{(11x+50)}{1}$$
$$(x-550)(11x+50)$$
A: Note that $27500 = 2500 \times 11$. Now divide $11x^2-6000x-27500$ by $2500$, the largest perfect square divisor of $27500$, which gives:
$$\frac{11x^2}{2500} - \frac{6000x}{2500} - 11 = 0$$
$$\Rightarrow 11 \left(\frac{x}{50}\right)^2 - \frac{6000}{50} \frac{x}{50} -11=0$$
$$\Rightarrow X = \frac{x}{50}: 11X^2 - 120X - 11 = 0$$
and since $11$ is prime, write this as $(11X + a)(X + b)$. Since $120$ is very close to $121 = 11^2$, $-120 = -11 \times 11 + 1$, so this factors as $(11X + 1)(X - 11) = 0 \Rightarrow X = -\frac{1}{11}, 11$. As $X = x/50$, $x = -\frac{50}{11}, 550$.
