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Solve for x: ​

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

Oct 17, 2019

#1
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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

$$\sqrt{1-4x^2}$$ _> 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
+108732
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Note: I have a=had to edit this a number of times becasue the LaTex compiler is playing up.

It keeps deleting things.

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

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