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

#1**+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

Melody
Mar 3, 2015

#1**+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

Melody
Mar 3, 2015