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Solve for x: \(\frac{\left(\sqrt{1-4x^2}-1\right)}{x}\ge -4\)

 Oct 17, 2019
 #1
avatar+999 
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I don't want to tell you the full answer, but I'll start you off.

Multiply both sides by x and get \(\sqrt{1-4x^2}-1 \geq -4x\).

To get rid of the radical, we can move all radical signs to the left and non-radicals to the right.

This is what we'll do here.

 

\(\sqrt{1-4x^2} \geq -4x+1\).

Now square both sides, and then you probably know what to do from there.

 

You are very welcome!

:P

 Oct 17, 2019
 #2
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what if x=0?

 Oct 17, 2019
 #3
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you can't multiply by variable!

 Oct 17, 2019
 #4
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You can. 

Guest Oct 18, 2019
 #5
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start by multiplying by x both sides to get 

\(\sqrt{1-4x^2}-1\) _> equal or more than, -4x 

Now add 1 

\(\sqrt{1-4x^2}\) _> equal or more than -4x+1

Square both sides

1-4x^2_> equal or more than (-4x+1)^2

subtract 1

-4x^2_>equal or more than (-4x+1)^2 -1 

divide by -4(Note we reverse the inequality sign) 

x^2_< equal or less than ((-4x+1)^2-1)/-4 (All divided by -4) 

now square root both sides

x _< equal or less than sqrt((-4x+1)^2-1)/-4))

 Oct 18, 2019
 #6
avatar+107006 
+1

Note: I have a=had to edit this a number of times becasue the LaTex compiler is playing up.

It keeps deleting things.  sad

 

\(\frac{\left(\sqrt{1-4x^2}-1\right)}{x}\ge -4\)

 

I will look at the domain restrictions first.

I am assuming that you want only real answers.

 

\(1-4x^2\ge0\\ 1\ge4x^2\\ 4x^2\le1\\ (2x)^2\le1\\ -1\le2x\le1\\ -0.5\le x\le0.5 \)

 

Oh, I canot divide by zero either so x cannot be zero

\(-0.5\le x<0 \qquad \)

or      \(0< x<0.5\)

Now that I have an initial restriction for x values I will look at the rest.

You cannot just multiply by x as Coolstuff has done because you do not know if x is positive or negative and if it is negative the sign has to be turned around.

Normally the easiest way to handle these is to multiply through by x^2 because we know that x^2 has to be positive. But in this case that will work well.

 

SO

When is

 \(\sqrt{1-4x^2}-1> 0\\ \sqrt{1-4x^2}>1\\ 1-4x^2>1\\ -4x^2>0\\ NO\;\;solutions.\)

 

If I had thought about it before hand I could have seen that this was the case.

So

\(\sqrt{1-4x^2}-1\le0\qquad \text{for all values of x}\)

 

If x is negative then the LHS of the original inequality will always be positive so the LHS will always be bigger than -4

 

So    all values of x where  \(-0.5\le x < 0\)   will make the equation true.

 

--------------------------------

If x is positive then

 

\(\frac{\left(\sqrt{1-4x^2}-1\right)}{x}\ge -4\\ \sqrt{1-4x^2}-1\ge -4x\\ \sqrt{1-4x^2}\ge -4x+1\\ \qquad If\;\;-4x+1<0\quad \text{then this must be true}\\ \qquad If\;\;-4x<-1\quad \text{then this must be true}\\ \qquad If\;\;x>-0.25\quad \text{then this must be true}\\ \qquad \text{Since I am only looking at positive values of x, x must be greater than -0.25}\)

 

So the solution set is     [-0.5, 0) union (0, 0.5]

 Oct 18, 2019
edited by Melody  Oct 18, 2019
edited by Melody  Oct 18, 2019

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