Solutions of an inequality (1-√(1-4x^2)/x < 3

[Vali]

I need to find the solutions of the following inequation:
(1-sqrt(1-4x^2)/x < 3
I put the conditions x different from 0 and 1-4x^2>=0 and I got [-1/2,0)U(0,1/2] which is the right answer but I'm confuse because I usually subtract 3 to get (1-sqrt(1-4x^2)/x - 3 < 0 then, after I made some work and I got a fraction, I find the variation of each function (from numerator and denominator) and in the final I find the sign of f(x) which should be negative in our case and like this I find the solutions, but I didn;t get the same result.

 

[HOI]

I need to find the solutions of the following inequation:
(1-sqrt(1-4x^2)/x < 3
I put the conditions x different from 0 and 1-4x^2>=0 and I got [-1/2,0)U(0,1/2] which is the right answer but I'm confuse because I usually subtract 3 to get (1-sqrt(1-4x^2)/x - 3 < 0 then, after I made some work and I got a fraction, I find the variation of each function (from numerator and denominator) and in the final I find the sign of f(x) which should be negative in our case and like this I find the solutions, but I didn;t get the same result.

You are missing a parenthesis. Do you mean (1- sqrt(1- 4x^2))/x< 3?

First, because of the $$\sqrt{1- 4x^2}$$, x must lie between -1/2 and 1/2.

In my opinion the best way to handle such an inequality is to first solve the associated equation, $$\frac{\sqrt{1- 4x^2}}{x}= 3$$. To solve that multiply both sides by x to get $$\sqrt{1- 4x^2}= 3x$$ and square both sides to get $$1- 4x^2= 9x^2$$. Then $$13x^2= 1$$, $$x= \pm\sqrt{1/13}$$.

The point of that is that in order that the value of a function change from "< 3" to "> 3" it must either be "= 3" or be discontinuous. This function is 3 at $$x=\pm\sqrt{1/13}$$ and, because of the division by 3, is discontinuous at x= 0.

Now, check one value in each interval, $$-1/2\le x\le -\sqrt{1/13}$$, $$-\sqrt{1/13}\le x< 0$$, $$0< x\le \sqrt{1/13}$$, and $$\sqrt{1/13}\le x\le 1/2$$ to determine in which interval the value is less than 3 and in which the value greater than 3.

 

[Vali]

Thank you for the help!
I understood, but I have one question.Why the associated equation is sqrt(1-4x^2)/x=3 and not (1-sqrt(1-4x^2))/x=3 ? Why you cancel that "1" ?

 

[HOI]

I just mistakenly dropped it! Yes, the associated equation should be $$\frac{1- \sqrt{1- x^2}}{x}= 3$$ so that $$1- \sqrt{1- 4x^2}= 3x$$. Then $$\sqrt{1- 4x^2}= 1- 3x$$ and, squaring both sides, $$1- 4x^2= 9x^2- 6x+ 1$$. That gives us the quadratic equation $$13x^2- 6x= x(13x- 6)= 0$$ which has roots 0 and -6/13.

 

[Wilmer]

That gives us the quadratic equation $$13x^2- 6x= x(13x- 6)= 0$$
which has roots 0 and -6/13.

0 and 6/13 :)