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\(\sqrt{a-\sqrt{a+x}}=x\)

Rollingblade  May 2, 2018
 #1
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Hey, Rollingblade, we meet again!

 

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
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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

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