Answer:
[tex]$x=\sqrt{\frac{7(4+\sqrt{15})}{2}} $[/tex]
Step-by-step explanation:
From the way it is written, the [tex]x[/tex] is outside the square root. I will rewrite it as:
[tex]x\sqrt{5} =x\sqrt{3} +\sqrt{7}[/tex]
[tex]x\sqrt{5}-x\sqrt{3}=\sqrt{7}[/tex]
[tex]x(\sqrt{5} - \sqrt{3} )=\sqrt{7}[/tex]
[tex]$x= \frac{\sqrt{7} }{\sqrt{5} - \sqrt{3}} \implies \frac{\sqrt{7}(\sqrt{5} + \sqrt{3}) }{2} $[/tex]
[tex]$x=\frac{1}{2} \sqrt{7} (\sqrt{5} + \sqrt{3} )$[/tex]
[tex]$x=\frac{\sqrt{35}}{2} +\frac{ \sqrt{21}}{2} $[/tex]
[tex]$x=\frac{\sqrt{35}+\sqrt{21}}{2} $[/tex]
Multiply denominator and numerator by 3
[tex]$x=\frac{3\sqrt{35}+3 \sqrt{21}}{6} $[/tex]
Factor [tex]\sqrt{3}[/tex]
[tex]\sqrt{3} (\sqrt{105}+3 \sqrt{7})[/tex]
[tex]$x=\frac{\sqrt{3} (\sqrt{105}+3 \sqrt{7})}{6} $[/tex]
Divide denominator and numerator by [tex]\sqrt{3}[/tex]
[tex]$x=\frac{\sqrt{105}+3 \sqrt{7}}{2\sqrt{3} } $[/tex]
Let's rewrite it again
[tex]$x=\frac{\sqrt{ (\sqrt{105}+3 \sqrt{7})^2}}{\sqrt{12} } $[/tex]
[tex]$x=\sqrt{ \frac{1}{12} \cdot (\sqrt{105}+3 \sqrt{7})^2}$[/tex]
It is already in the form [tex]$\sqrt{\frac{a}{b} } $[/tex]
Expanding the perfect square, we have
[tex]63+42\sqrt{15}+105[/tex]
[tex]$\frac{63}{12} +\frac{42\sqrt{15}}{12} +\frac{105}{12} $[/tex]
[tex]$\frac{21}{4} +\frac{7\sqrt{15}}{2} +\frac{35}{4} $[/tex]
Factor [tex]$\frac{7}{2} $[/tex]
[tex]$\frac{7}{2} (4+\sqrt{15} )$[/tex]
Therefore,
[tex]$x=\sqrt{\frac{7}{2} \left(4+\sqrt{15} \right)} $[/tex]
[tex]$x=\sqrt{\frac{7(4+\sqrt{15})}{2}} $[/tex]
