# 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.

• I hope the new title reflects what you hope to obtain in an answer. If not, please let me know. – Lord_Farin Jun 7 '13 at 13:14
• Note that the two roots aren't factors of 27500 and don't sum to 6000. Here since the leading coefficient is not 1, their product and sum respectively have to be 27500/11 and 6000/11. – Milind Jun 7 '13 at 13:14
• The two roots are $x_1\approx270$ and $x_1\approx275$. I obtained this using the quadratic formula but without using a calculator. Does it count? – Andrea Mori Jun 7 '13 at 13:27
• No, the roots are $x_1=550$ and $x_2=-50/11$. WolframAlpha confirms. – Milind Jun 7 '13 at 13:29
• Right, I copied the equation wrong on a piece of paper! :) Roots are indeed those, and can be actually computed by hand, no calculator needed. – Andrea Mori Jun 7 '13 at 13:35

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 $-27500/11$.

• awesome. but how u figured its 500 and 600 [i'm a beginner i said.] – Dineshkumar Jun 7 '13 at 13:25
• Ah, that step is one of the reasons I called this an ad-hoc method. I personally noticed that 5 and 6 are consecutive factors of 30. – Milind Jun 7 '13 at 13:27
• How about -50/11 ? – Dineshkumar Jun 7 '13 at 13:35
• @MilindHegde : also $150$ and $250$ are two integers with difference $100$ dividing $3000$. Also $50$ and $150$. Also $-50$ and $50$ ... – Andrea Mori Jun 7 '13 at 13:47
• That's true. My mind tried multiples of 100 first. It is also a good idea too see approximately how big $3000/y$ is to quickly exclude possibilities like 50 and 100. – Milind Jun 7 '13 at 13:51

## 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...

• explanation for 50(y-1) and 50(y+1) please. – Dineshkumar Jun 7 '13 at 13:38
• @Dineshkumar $x-50 = 50y-50 = 50(y-1)$. – Kaster Jun 7 '13 at 14:43
• @JoelReyesNoche Thanks. Quick typo. – Kaster Jun 7 '13 at 14:44

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

$$(y+3000)^2=3050^2$$

• Nice way to remove the annoying 11 ! – Frédéric Grosshans Jun 7 '13 at 15:51
• By the way, on the line you call "last", I find the following development, which completes the square, more intuitive $3000^2+ 302500= 3000^2+ 3025\times100= 3000^2 + 3000\times100+25\times100 = 3000^2+2\times3000\times50+50^2=3050^2$ – Frédéric Grosshans Jun 7 '13 at 15:56
• The last line should read $(y - 3000)^2 = 3050^2$ – Happy Green Kid Naps Jun 7 '13 at 21:39

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:

$$11x^2-6000x-27500=0$$ $$11x^2-6050x+50x-27500=0$$ $$11x(x-550)+50(x-550)=0$$ $$(x-550)(11x+50)=0$$ $$(x-550)=0\;\;,(11x+50)=0$$ $$x=550,-\dfrac{50}{11}$$

• I tried but getting big value and some errors. – Dineshkumar Jun 7 '13 at 13:18
• @Dineshkumar I think you face problem to solve square root of $37210000$. – iostream007 Jun 7 '13 at 13:22
• How u added and subracted 50? bringing it 11x(x−550)+50(x−550)=0? i'm a beginner! – Dineshkumar Jun 7 '13 at 13:27
• using second approach is a guess for big numbers.I want to part 6000 in such a way that multiplication of those part will be -302500.since multiplication sign is - so piece of 6000 is one is bigger than 6000 and one is less than.so just take some easy value and check their multiplication if it is near to your desired value than just little bit change the value.I solve it i n 3 times firstly I took 6500 and 500,then 6100,100 then 6050 and 50 – iostream007 Jun 7 '13 at 13:32
• @Beska I don't know either, but $60^2=3600$ obviously, and it's not far from $3721$ from below, so $61^2$ is a natural guess because of the final digit $1$. Now, $61^2=(60+1)^2=60^2+2\cdot60\cdot1+1^2=3600+120+1=3721$ ... Bingo! ... All of this can easily be done in your head. – Andrea Mori Jun 7 '13 at 16:23

$$11x^2 - 6000x - 27500 = 0$$

$a=11$

$b=-6000 = 2^4 \cdot 3 \cdot 5^3$

$c = -27500 = 2^2 \cdot 5^4 \cdot 11$

$ac=-2^2 \cdot 5^4 \cdot 11^2$

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

$$\dfrac{(11x-6050)}{11} \dfrac{(11x+50)}{1}$$

$$(x-550)(11x+50)$$