Question

2 - sqrt (3x - 1) - 1 / (a^2 - x^2) = 0
Find the range of values of a for which the equation has 2 real roots.

Answer 1

oh my friend @TURITW I think your question is really interesting, I tried many times to do it algebraically but failed every time, so my solution will be concluded by studying some special cases.
okay we start by noticing the term \[\frac{ 1 }{a^2- x^2 }\]
we have a vertical asypmtote at x=|a| , also we notice that \[\lim_{x \rightarrow a+}f(x)=+\infty ,\lim_{x \rightarrow a-}=-\infty\] the function is continuous at x>=(1/3) at evry value but a ,so by the intermediate value theorem there must be at least one root for f(x) for every value to x>=(1/3).
now we evaluate f(1/3)=0 to see the minimum value that a can take, and I chose f(1/3)=0 because (1/3) is the first number in the domain so when f(1/3)=0 we can say that there are only two roots according to the behaviour of f(x) that I just told you, so:\[f(\frac{ 1 }{ 3 })=2-0-\frac{ 1 }{ a^2-\frac{ 1 }{ 9 } }=0\]\[a=\pm\frac{ 11 }{ 18 }\]
so after all this we can finally say that |a|>(11/18) , which means\[a\in (-\infty,\ -\sqrt{\frac{ 11 }{ 18 }}] \cup [\sqrt{\frac{ 11 }{ 18 }},+\infty)\]

Answer 2

@ybarrap have you seen something like this before? :\

Answer 3

can you find it using an algebraic solution, or it's not possible?

Answer 4

Is this the problem?
$$
2 - sqrt (3x - 1) - \dfrac{1} {a^2 - x^2} = 0
$$

Answer 5

yes

Answer 6

Oh thanks, the answer you got is the same as the answer given.

Answer 7

that's amazing then :), you are so welcome @TURITW