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

Vali

Member
Dec 29, 2018
48
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.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
368
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 [tex]\sqrt{1- 4x^2}[/tex], 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, [tex]\frac{\sqrt{1- 4x^2}}{x}= 3[/tex]. To solve that multiply both sides by x to get [tex]\sqrt{1- 4x^2}= 3x[/tex] and square both sides to get [tex]1- 4x^2= 9x^2[/tex]. Then [tex]13x^2= 1[/tex], [tex]x= \pm\sqrt{1/13}[/tex].

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 [tex]x=\pm\sqrt{1/13}[/tex] and, because of the division by 3, is discontinuous at x= 0.

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

Vali

Member
Dec 29, 2018
48
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" ?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
368
I just mistakenly dropped it! Yes, the associated equation should be [tex]\frac{1- \sqrt{1- x^2}}{x}= 3[/tex] so that [tex]1- \sqrt{1- 4x^2}= 3x[/tex]. Then [tex]\sqrt{1- 4x^2}= 1- 3x[/tex] and, squaring both sides, [tex]1- 4x^2= 9x^2- 6x+ 1[/tex]. That gives us the quadratic equation [tex]13x^2- 6x= x(13x- 6)= 0[/tex] which has roots 0 and -6/13.
 

Wilmer

In Memoriam
Mar 19, 2012
376