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# I would like to know the solution of the following equation.

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

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

Hamed

Mar 2, 2015

#1
+95360
+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

Mar 3, 2015

#1
+95360
+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
#2
+95360
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

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

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Mar 3, 2015