I would like to know the solution of the following equation:
$$\sqrt{a-\sqrt{a+x}} = x$$
Thanks in advance,
Hamed
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
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