Problema
(Indicado a partir do 9º ano do E. F.)
Dados os números reais [tex]a[/tex] e [tex]b[/tex], onde
[tex]\qquad a^2+b^2=\left(\dfrac{56}{13}\right)^2[/tex] e [tex]a+\dfrac{5b}{12}=\dfrac{56}{12},[/tex]
determine o valor de [tex]a+b[/tex].
Solução
De [tex]a+\dfrac{5b}{12}=\dfrac{56}{12}[/tex], vem [tex]a=\dfrac{56-5b}{12}.[/tex]
Daí,
[tex]\qquad a^2+b^2=\left(\dfrac{56}{13}\right)^2[/tex]
[tex]\qquad \left(\dfrac{56-5b}{12}\right)^2+b^2=\dfrac{56^2}{13^2}[/tex]
[tex]\qquad \left(\dfrac{56}{12}\right)^2-\dfrac{560b}{144}+\left(\dfrac{5b}{12}\right)^2+b^2=\dfrac{56^2}{13^2}[/tex]
[tex]\qquad \left(\dfrac{14}{3}\right)^2-\dfrac{35b}{9}+\left(\dfrac{5b}{12}\right)^2+b^2=\dfrac{56^2}{13^2}[/tex]
[tex]\qquad \dfrac{196}{9}-\dfrac{35b}{9}+\dfrac{25b^2}{144}+b^2=\dfrac{3136}{169}[/tex]
[tex]\qquad \dfrac{196}{9}-\dfrac{35b}{9}+\dfrac{169b^2}{144}=\dfrac{3136}{169}[/tex]
[tex]\qquad \dfrac{169b^2}{144}-\dfrac{35b}{9}+\dfrac{4900}{1521}=0[/tex]
[tex]\qquad 169b^2-560b+\dfrac{78400}{169}=0[/tex]
[tex]\qquad b = \dfrac{560\pm\sqrt{560^2-4\cdot \cancel{169}\cdot \frac{78400}{\cancel{169}}}}{2\cdot 169}[/tex]
[tex]\qquad b = \dfrac{560\pm\sqrt{560^2-4\cdot 78400}}{2\cdot 169}[/tex]
[tex]\qquad b = \dfrac{560\pm\sqrt{0}}{2\cdot 169}[/tex]
[tex]\qquad b = \dfrac{560}{2\cdot 169}[/tex]
[tex]\qquad b = \dfrac{280}{169}.[/tex]
Assim,
[tex]\qquad a=\dfrac{56-5b}{12}=\dfrac{56-5\cdot \frac{280}{169}}{12}=\dfrac{\frac{8064}{169}}{12}=\dfrac{672}{169}.[/tex]
Finalmente,
[tex]\qquad a+b=\dfrac{672}{169}+\dfrac{280}{169}=\dfrac{952}{169}[/tex].
Solução elaborada pelos Moderadores do Blog.