.Desafio: $\alpha = \beta ?$

Problema
(Indicado a partir do 9º ano do E. F.)

$\qquad \alpha=\sqrt{13}+\sqrt{10+2\sqrt{13}}$
e
$\qquad \beta=\sqrt{5+2\sqrt{3}}+\sqrt{18-2\sqrt{3}+2\sqrt{65-26\sqrt{3}}}$ .

Mostre que $\alpha=\beta$.

Solução

• Inicialmente, observe que

  $\qquad\qquad\beta=\sqrt{5+2\sqrt{3}}+\sqrt{18-2\sqrt{3}+2\sqrt{65-26\sqrt{3}}}$

  $\qquad\qquad\beta=\sqrt{5+2\sqrt{3}}+\sqrt{18-2\sqrt{3}+2\sqrt{13}\sqrt{5-2\sqrt{3}}}$

  $\qquad\qquad\beta=\sqrt{5+2\sqrt{3}}+\sqrt{\left(\sqrt{13}+\sqrt{5-2\sqrt{3}}\right)^2}$

  $\qquad\qquad\boxed{\beta=\sqrt{5+2\sqrt{3}}+\sqrt{13}+\sqrt{5-2\sqrt{3}}}$.

• Agora, seja $\boxed{x=\sqrt{5+2\sqrt{3}}+\sqrt{5-2\sqrt{3}}}$.
  Assim:
  $\qquad\qquad x^2=5+2\sqrt{3}+2\sqrt{5+2\sqrt{3}}\sqrt{5-2\sqrt{3}}+5-2\sqrt{3}$

  $\qquad\qquad x^2=10+2\sqrt{25-12}$

  $\qquad\qquad x^2=10+2\sqrt{13}$

  $\qquad\qquad x=\pm\sqrt{10+2\sqrt{13}}.$

  Como $x\gt 0$, então $\boxed{x=\sqrt{10+2\sqrt{13}}}$.

Portanto, pelo exposto,

$\qquad \qquad \alpha=\sqrt{13}+\sqrt{10+2\sqrt{13}}$

$\qquad \qquad \alpha=\sqrt{13}+x$

$\qquad \qquad \alpha=\sqrt{13}+\left[\sqrt{5+2\sqrt{3}}+\sqrt{5-2\sqrt{3}}\right]$

$\qquad \qquad \alpha=\sqrt{5+2\sqrt{3}}+\sqrt{13}+\sqrt{5-2\sqrt{3}}$

$\qquad\qquad\alpha=\beta$.

Solução elaborada pelos Moderadores do Blog.