The value under the square root can't be negative. This affects the domain: in addition to x != 0, we need 1-4x^2 to be non-negative. The absolute value you are asking about is the solution of this inequality.Observe first picture to see how I attempted to solve this problem. Observe the second picture to see how the solution shows it done. Where did the absolute come from along with everything else that is not like how I thought it should be? I don't even know where I went wrong...
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Okay, thank you.The value under the square root can't be negative. This affects the domain: in addition to x != 0, we need 1-4x^2 to be non-negative. The absolute value you are asking about is the solution of this inequality.
Edit: Just looked at your solution - you got the same result, just a different method.
Where's that?I must be missing something here: for 0.5< x<0 the left hand side is negative, so how can it be > 3/2 ?
The first line from notebook says [imath] \frac{1-\sqrt{1-4x^2}{x} > \frac{3}{2}[/imath]. When [imath]x[/imath] is negative the whole left hand side is negative too. (For some reason the math in this reply did not typeset correctly, no clue why).Where's that?
And in your notes you seem to have multiplied both sides by x, which only preserves inequalities when x is positive.The first line from notebook says [imath] \frac{1-\sqrt{1-4x^2}{x} > \frac{3}{2}[/imath]. When [imath]x[/imath] is negative the whole left hand side is negative too. (For some reason the math in this reply did not typeset correctly, no clue why).
I didn't multiple anything by x.And in your notes you seem to have multiplied both sides by x, which only preserves inequalities when x is positive.
You had to multiply by x to get from x1−1−4x2>23 to 1−4x2<1−23xI didn't multiple anything by x.
It did not render because it is in error. You need to enclose the radical in a closing curly bracket and then enclose the numerator in a second closing curly bracket. You had only one.The first line from notebook says [imath] \frac{1-\sqrt{1-4x^2}{x} > \frac{3}{2}[/imath]. When [imath]x[/imath] is negative the whole left hand side is negative too. (For some reason the math in this reply did not typeset correctly, no clue why).
No. I don’t think you did multiply by x. Rather you multiplied by 4x^2, which entailed multiplying by a positive number. But earlier you squared fractions, which raises the same issue in an even more complex way.I didn't multiple anything by x.
Think again.I didn't multiple anything by x.