MHB Solve Equation: $\sqrt{1+\sqrt{1-x^2}}$

  • Thread starter Thread starter anemone
  • Start date Start date
Click For Summary
The equation $\sqrt{1+\sqrt{1-x^2}}(\sqrt{(1+x)^3}-\sqrt{(1-x)^3})=2+\sqrt{1-x^2}$ was discussed, leading to the conclusion that the only solution is $x=\dfrac{1}{\sqrt{2}}$. The participants confirmed this result with agreement. The interaction was friendly, with expressions of gratitude and acknowledgment of correctness. This solution highlights the specific value of $x$ that satisfies the equation. The discussion effectively resolved the problem presented.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Solve the equation $\sqrt{1+\sqrt{1-x^2}}(\sqrt{(1+x)^3}-\sqrt{(1-x)^3}=2+\sqrt{1-x^2}$.
 
Mathematics news on Phys.org
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 $x = \pm\frac{1}{\sqrt{2}}$.
 
Hi lfdahl! The correct answer is $x=\dfrac{1}{\sqrt{2}}$ only. (Smile)
 
Thankyou, anemone - of course you're right!👍