can i solve this quadratic equation this way

Question

I was basically doing a physics problem and came across this equation in midway

$\dfrac{2n-1}{n^2} = \dfrac{11}{36}$

then I equated $2n-1 = 11$ and $n^2 = 36$ and the value of $n$ which I got is $6$ in both the cases

is this method valid? I mean I equated the ratio and ratio between many numbers are same

I just want to know that whether I did it correctly or not.

@DavidH it can be made into a quadratic equation just by cross mutliplication — agha rehan abbas, Jul 12 '14 at 05:39
You got lucky. The method will in general not work. And even in this case there is another (negative) root. — André Nicolas, Jul 12 '14 at 05:43
I was referring to the equation prior to edit, which was a cubic. — David H, Jul 12 '14 at 05:43
No, as you miss the other solution that exists for this problem. — JB King, Jul 12 '14 at 05:47

score 3 · Accepted Answer · answered Jul 12 '14 at 05:45

3

As a proportion: $\dfrac{A}{B} = \dfrac{C}{D} \to AD = BC$ we have:

$36(2n-1) = 11n^2$. So: $11n^2 - 72n + 36 = 0$, and $(11n-6)(n-6) = 0$. So $n = 6$ and $n = \dfrac{6}{11}$

answered Jul 12 '14 at 05:45

DeepSea

77,651

thanks it helped me a lot i was not able to solve the equation – agha rehan abbas Jul 12 '14 at 05:49

score 1 · Answer 2 · answered Jul 12 '14 at 05:50

1

No, it's not valid. For example, you can't solve $\dfrac{2n - 1}{n^2} = \dfrac{22}{36}$ in this way, but it's mathematically the same equation.

And before you think it has anything to do with the "simplest form" of the fraction on the RHS, consider $\dfrac{2n - 3}{n^2} = \dfrac{1}{3}$. The actual solution is $n = 3$, but by equating $2n - 3$ to $1$, you get $n = 2$.

answered Jul 12 '14 at 05:50

M. Vinay

9,004

i think you misread my equation it is 2n-1 and not 2n-3 – agha rehan abbas Jul 12 '14 at 05:52
1

@agharehanabbas I was giving another example that is similar (but not the same, obviously). Your answer is, of course, correct, but your question was about the validity of the method you used. I was attempting to answer that by showing an example where it does not give the correct solution. – M. Vinay Jul 12 '14 at 06:38
ok now i got your point of view – agha rehan abbas Jul 12 '14 at 06:40

score 0 · Answer 3 · answered Jul 12 '14 at 06:16

It was by pure dumb luck that you were able to obtain a solution "by inspection". Your solution is provably correct: if you substitute $n=6$, you do indeed get 11 for the numerator and 36 for the denominator.

However, the technique only worked in this case because of the exceptional coincidence that the same guess for $n$ happened to make everything work. In other words, the technique doesn't generalize at all, and thus most people wouldn't have even noticed that a solution could be obtained by inspection.

Also, every quadratic has up to two real roots; I don't see any shortcut for obtaining the other real root.

can i solve this quadratic equation this way

3 Answers3