Square root of an irrational number

We note:-
$\sqrt{\sqrt{a}}*\sqrt{b}=\sqrt{\sqrt{a}}*\sqrt{\sqrt{b^2}} =\sqrt{\sqrt{a*b^2}} \\$
$\sqrt{48}-\sqrt{45}=\sqrt3*(\sqrt{16}-\sqrt{15}) \\=\sqrt3*(4-\sqrt{15}). \\4-\sqrt{15}=(\dfrac{5}{2}+\dfrac{3}{2}-2*\sqrt{\dfrac{5}{2}}*\sqrt{\dfrac{3}{2}})\\ =(\sqrt{\dfrac{5}{2}}-\sqrt{\dfrac{3}{2}})^2 \\ \therefore~ \sqrt{\sqrt{48}-\sqrt{45}} \\ =\sqrt{\sqrt3}*\sqrt{(\sqrt{ \dfrac{5}{2} }-\sqrt{ \dfrac{3}{2} } )^2} \\=\sqrt{\sqrt3}*(\sqrt{ \dfrac{5}{2}}-\sqrt{\dfrac{3}{2} }) \\= \sqrt{\sqrt3}*(\sqrt{\sqrt{ \dfrac{25}{4}}}-\sqrt{\sqrt{\dfrac{9}{4} } }) \\=\sqrt{\sqrt{ \dfrac{75}{4}}}-\sqrt{\sqrt{\dfrac{27}{4} } }$
In algebra, the best first step is to factor out a common factor from an expression. In the above it was $\sqrt3$ . Say we have $12x^2 - 27$ unless we factor out 3, it will not be possible to do further factors.,

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