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I would like to know the solution of the following equation:

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

Thanks in advance,

Hamed

Guest Mar 2, 2015

Best Answer 

 #1
avatar+93616 
+5

What do you mean by solution.  there are 2 unknowns  Do you want the integer solutions?

I'll try to make x the subject

 

$$\begin{array}{rll}
\sqrt{a-\sqrt{a-x}}&=&x\\\\
a-\sqrt{a-x}&=&x^2\\\\
-\sqrt{a-x}&=&x^2-a\\\\
a-x&=&(x^2-a)^2\\\\
a-x&=&x^4-2ax^2+a^2\\\\
a-a^2-x+2ax^2-x^4&=&0\\\\
-((a-x-x^2)(-1+a+x-x^2))&=&0\\\\
(a-x-x^2)(-1+a+x-x^2)&=&0\\\\
(a-x-x^2)(-1+a+x-x^2)&=&0\\\\
\end{array}
so \\
a-x-x^2=0\qquad or \qquad -1+a+x-x^2=0\\\\
x^2+x-a=0\qquad or \qquad x^2-x+(1-a)=0\\\\$$

 

You can now solve these seperately using the quadratic formula and you will end up with 4 solutions where x is the subject.

 

the numeric value of x will depend on your value of a.

here is a graph

https://www.desmos.com/calculator/oxyumtk0dq

Melody  Mar 3, 2015
 #1
avatar+93616 
+5
Best Answer

What do you mean by solution.  there are 2 unknowns  Do you want the integer solutions?

I'll try to make x the subject

 

$$\begin{array}{rll}
\sqrt{a-\sqrt{a-x}}&=&x\\\\
a-\sqrt{a-x}&=&x^2\\\\
-\sqrt{a-x}&=&x^2-a\\\\
a-x&=&(x^2-a)^2\\\\
a-x&=&x^4-2ax^2+a^2\\\\
a-a^2-x+2ax^2-x^4&=&0\\\\
-((a-x-x^2)(-1+a+x-x^2))&=&0\\\\
(a-x-x^2)(-1+a+x-x^2)&=&0\\\\
(a-x-x^2)(-1+a+x-x^2)&=&0\\\\
\end{array}
so \\
a-x-x^2=0\qquad or \qquad -1+a+x-x^2=0\\\\
x^2+x-a=0\qquad or \qquad x^2-x+(1-a)=0\\\\$$

 

You can now solve these seperately using the quadratic formula and you will end up with 4 solutions where x is the subject.

 

the numeric value of x will depend on your value of a.

here is a graph

https://www.desmos.com/calculator/oxyumtk0dq

Melody  Mar 3, 2015
 #2
avatar+93616 
0

I cheated a little by getting Wolfram Alpha to do this factorising step for me.

Can anyone show me how to do it by hand?    Thank you  

 

 

$$a-a^2-x+2ax^2-x^4&=&0\\\\
-((a-x-x^2)(-1+a+x-x^2))&=&0\\\\$$

Melody  Mar 3, 2015

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