Yes, it satisfies Pythagoras equation: \displaystyle{\left(\sqrt{{{12}}}\right)}^{{2}}={\left(\sqrt{{{5}}}\right)}^{{2}}+{\left(\sqrt{{{7}}}\right)}^{{2}} So this is a right angled triangle with ...

\displaystyle{\left({4}{\left({3}\sqrt{{3}}-{4}\right)}\right.} Explanation: \displaystyle{\left(-{2}\sqrt{{3}}+{2}\right)}{\left(\sqrt{{3}}-{5}\right)} \displaystyle\therefore\sqrt{{3}}\times\sqrt{{3}}={3} ...

See a solution process below: Explanation: First, expand the terms in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis: \displaystyle{\left({2}\sqrt{{{3}}}\right)}{\left({2}\sqrt{{{3}}}-{4}\sqrt{{{5}}}\right)}\Rightarrow ...

See a solution process below: Explanation: First, we can rewrite the expression as: \displaystyle{2}{\sqrt[{{5}}]{{{2}}}}+{2}{\sqrt[{{5}}]{{{32}\cdot{2}}}}\Rightarrow \displaystyle{2}{\sqrt[{{5}}]{{{2}}}}+{2}{\sqrt[{{5}}]{{{2}^{{5}}\cdot{2}}}} ...

One can use the fact that \mathbb{Q}(\sqrt2,\sqrt3)=(\mathbb{Q}(\sqrt2))(\sqrt3) the latter of which has elements of the form a+b\sqrt3 where a,b\in \mathbb{Q}(\sqrt2) since [(\mathbb{Q}(\sqrt2))(\sqrt3) : \mathbb Q(\sqrt2)]=2 ...

To be able to compute the roots of 3^{2x} + k3^x +2 with respect to 3^x you have immediately the condition k^2 - 8 \geq 0. So k\leq -2\sqrt{2} or k\geq 2\sqrt{2}. In case of either equality, ...