Question

[tex] \sqrt{24-x^2}-\sqrt{8-x^2}=2

\sqrt{24-x^2}+\sqrt{8-x^2}=? [/tex]

e de liceu

Answer 1

[tex]\text{Amplificam cu conjugatul prima ecuatie obtinem:}\\ \dfrac{(\sqrt{24-x^2}-\sqrt{8-x^2})(\sqrt{24-x^2}+\sqrt{8-x^2})}{\sqrt{24-x^2}+\sqrt{8-x^2}} = \dfrac{24-x^2-8+x^2}{\sqrt{24-x^2}+\sqrt{8-x^2}}=\\ =\dfrac{16}{\sqrt{24-x^2}+\sqrt{8-x^2}}=2\\ \text{Deci } \sqrt{24-x^2}+\sqrt{8-x^2}=\dfrac{16}{2}=\boxed{8}[/tex]

Answer 2

$\sqrt{24-x^2}-\sqrt{8-x^2} = 2\\ \\ u = \sqrt{24-x^2},\quad v = \sqrt{8 - x^2} \\ \\ u^2-v^2 = (u+v)(u-v)\\ 24-x^2-(8-x^2) = (u+v)(u-v) \\ 16 = (u+v)(u-v) \\ 16 = (u+v)\cdot 2 \\ \Rightarrow u+v = \dfrac{16}{2} \Rightarrow u+v = 8 \Rightarrow \sqrt{24-x^2}+\sqrt{8-x^2} = 8$