Question

$$\frac{1}{8(1-\sqrt{x})=\frac{1}{8(1+\sqrt{x})}+\frac{2\sqrt{x}}{8(1-\sqrt{x})}$$

Answer

x=0

Solution


Simplify  \(2\sqrt{x}\times 8(1-\sqrt{x})\)  to  \(2\times 8\sqrt{x}(1-\sqrt{x})\).
\[18(1-\sqrt{x})=18(1+\sqrt{x})+2\times 8\sqrt{x}(1-\sqrt{x})\]
Simplify  \(18(1+\sqrt{x})+2\times 8\sqrt{x}(1-\sqrt{x})\)  to  \(18(1+\sqrt{x})+16\sqrt{x}(1-\sqrt{x})\).
\[18(1-\sqrt{x})=18(1+\sqrt{x})+16\sqrt{x}(1-\sqrt{x})\]
Separate terms with roots from terms without roots.
\[18(1-\sqrt{x})-18(1+\sqrt{x})=16\sqrt{x}(1-\sqrt{x})\]
Square both sides.
\[1296x=256x{(1-\sqrt{x})}^{2}\]
Expand.
\[1296x=256x-512{x}^{\frac{3}{2}}+256{x}^{2}\]
Subtract \(256x\) from both sides.
\[1296x-256x=-512{x}^{\frac{3}{2}}+256{x}^{2}\]
Simplify  \(1296x-256x\)  to  \(1040x\).
\[1040x=-512{x}^{\frac{3}{2}}+256{x}^{2}\]
Move all terms to one side.
\[1040x+512{x}^{\frac{3}{2}}-256{x}^{2}=0\]
Factor out the common term \(16\).
\[16(65x+32{x}^{\frac{3}{2}}-16{x}^{2})=0\]
Divide both sides by \(16\).
\[65x+32{x}^{\frac{3}{2}}-16{x}^{2}=0\]
No root was found algebraically. However, the following root(s) were found by numerical methods.
\[x=0\]