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help, so hard!

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$$\sqrt{a-\sqrt{a+x}}=x$$

#1
+867
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Here is my solution from the same problem I posted.

We can designate $$\sqrt{a+x}=y$$, which means that:

$$\sqrt{a-y}=x$$

From this system, we can square both.

$$a+x=y^2$$ and $$a-y=x^2$$

When we subtract the second equation from the first, we get:

$$x+y=y^2-x^2$$

This also is:

$$x^2-y^2+x+y=(x+y)(x-y+1)=0$$

From this, we have two cases:

Case 1: x + y = 0

In this case, y = -x and $$x^2-x-a=0$$

Case 2: y = x + 1 and $$x^2+x+1-a=0$$

Solving for x in both equations, we get.

$$x_{1,2}=\frac{1}{2}\pm\sqrt{a+\frac{1}{4}}$$

And

$$x_{3,4}=-\frac{1}{2}\pm\sqrt{a-\frac{3}{4}}$$

I hope this helped,

Gavin

GYanggg  May 2, 2018
edited by GYanggg  Jun 21, 2018
#2
0

1. Did you mean (a-x)1/2=y?

2. Do x and a have to be real? (There are no real solutions to this equation)

Guest May 3, 2018
#3
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Nvm there are real solutions to this equation.

Sorry!

Guest May 3, 2018