Question

If

Answer 1

Answer:

$x=\sqrt{\frac{7(4+\sqrt{15})}{2}}$

Step-by-step explanation:

From the way it is written, the $x$ is outside the square root. I will rewrite it as:

$x\sqrt{5} =x\sqrt{3} +\sqrt{7}$

$x\sqrt{5}-x\sqrt{3}=\sqrt{7}$

$x(\sqrt{5} - \sqrt{3} )=\sqrt{7}$

$x= \frac{\sqrt{7} }{\sqrt{5} - \sqrt{3}} \implies \frac{\sqrt{7}(\sqrt{5} + \sqrt{3}) }{2}$

$x=\frac{1}{2} \sqrt{7} (\sqrt{5} + \sqrt{3} )$

$x=\frac{\sqrt{35}}{2} +\frac{ \sqrt{21}}{2}$

$x=\frac{\sqrt{35}+\sqrt{21}}{2}$

Multiply denominator and numerator by 3

$x=\frac{3\sqrt{35}+3 \sqrt{21}}{6}$

Factor $\sqrt{3}$

$\sqrt{3} (\sqrt{105}+3 \sqrt{7})$

$x=\frac{\sqrt{3} (\sqrt{105}+3 \sqrt{7})}{6}$

Divide denominator and numerator by $\sqrt{3}$

$x=\frac{\sqrt{105}+3 \sqrt{7}}{2\sqrt{3} }$

Let's rewrite it again

$x=\frac{\sqrt{ (\sqrt{105}+3 \sqrt{7})^2}}{\sqrt{12} }$

$x=\sqrt{ \frac{1}{12} \cdot (\sqrt{105}+3 \sqrt{7})^2}$

It is already in the form $\sqrt{\frac{a}{b} }$

Expanding the perfect square, we have

$63+42\sqrt{15}+105$

$\frac{63}{12} +\frac{42\sqrt{15}}{12} +\frac{105}{12}$

$\frac{21}{4} +\frac{7\sqrt{15}}{2} +\frac{35}{4}$

Factor $\frac{7}{2}$

$\frac{7}{2} (4+\sqrt{15} )$

Therefore,

$x=\sqrt{\frac{7}{2} \left(4+\sqrt{15} \right)}$

$x=\sqrt{\frac{7(4+\sqrt{15})}{2}}$

Answer 2

X= 7/4 x 25/6 is the answer