Sep 25, 2020 Thread starter Admin #1 anemone MHB POTW Director Staff member Feb 14, 2012 3,812 Solve the equation $\sqrt{1+\sqrt{1-x^2}}(\sqrt{(1+x)^3}-\sqrt{(1-x)^3}=2+\sqrt{1-x^2}$.
Sep 29, 2020 #2 lfdahl Well-known member Nov 26, 2013 733 Spoiler: Suggested solution Rewriting the equation with $a = \sqrt{1+x}$ and $b = \sqrt{1-x}$: $\sqrt{1+ab}\left ( a^3 - b^3 \right ) = 2 + ab$ $\sqrt{1+ab}\left ( a - b \right )(a^2+b^2+ab) = 2 + ab$ $\sqrt{1+ab}\left ( a - b \right ) = 1$ - because $a^2+b^2 = 2$. Squaring yields: $(1+ab)\left ( a^2 + b^2-2ab \right ) = 1$ or $2(1+ab)(1-ab)= 1$ or $(ab)^2 = \frac{1}{2}$ hence $1-x^2 =\frac{1}{2}$ or $x = \pm\frac{1}{\sqrt{2}}$.
Spoiler: Suggested solution Rewriting the equation with $a = \sqrt{1+x}$ and $b = \sqrt{1-x}$: $\sqrt{1+ab}\left ( a^3 - b^3 \right ) = 2 + ab$ $\sqrt{1+ab}\left ( a - b \right )(a^2+b^2+ab) = 2 + ab$ $\sqrt{1+ab}\left ( a - b \right ) = 1$ - because $a^2+b^2 = 2$. Squaring yields: $(1+ab)\left ( a^2 + b^2-2ab \right ) = 1$ or $2(1+ab)(1-ab)= 1$ or $(ab)^2 = \frac{1}{2}$ hence $1-x^2 =\frac{1}{2}$ or $x = \pm\frac{1}{\sqrt{2}}$.
Sep 29, 2020 Thread starter Admin #3 anemone MHB POTW Director Staff member Feb 14, 2012 3,812 Hi lfdahl ! The correct answer is $x=\dfrac{1}{\sqrt{2}}$ only.
Sep 29, 2020 #4 lfdahl Well-known member Nov 26, 2013 733 Thankyou, anemone - of course you're right!