Question

\(\sqrt{4-\sqrt{7}}\cdot\sqrt{4+\sqrt{7}}=\)

Answer 1

`\sqrt(4-\sqrt7). \sqrt(4+\sqrt7) =\sqrt((4-\sqrt7)(4+\sqrt7))=\sqrt(4^2-\sqrt7^2)=\sqrt(16-7)=\sqrt9=3`

Answer 2

\(\sqrt{4-\sqrt{7}}\cdot\sqrt{4+\sqrt{7}}\)

\(=\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)

=3

Answer 3

\(\sqrt{4-\sqrt{7}}.\sqrt{4+\sqrt{7}}=\sqrt{\dfrac{8-2\sqrt{7}}{2}}.\sqrt{\dfrac{8+2\sqrt{7}}{2}}\)

                                 \(=\sqrt{\dfrac{7-2\sqrt{7}+1}{2}}.\sqrt{\dfrac{7+2\sqrt{7}+1}{2}}\)

                                 \(=\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}.\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}\)

                                 \(=\dfrac{\sqrt{7}-1}{\sqrt{2}}.\dfrac{\sqrt{7}+1}{\sqrt{2}}=\dfrac{7-1}{2}=3\)