Question

Se $a ^{1/2} +a ^{-1/2} = \frac{10}{3}$, então a+a vale:

Answer 1

$a^{1/2}+a^{-1/2}=\dfrac{10}{3}$

Elevando os dois lados da equação ao quadrado:


$(a^{\frac{1}{2}}+a^{-\frac{1}{2}})^{2}=\left(\dfrac{10}{3}\right)^{2}\\\\\\(a^{\frac{1}{2}})^{2}+2\cdot a^{\frac{1}{2}}\cdot a^{-\frac{1}{2}}+(a^{-\frac{1}{2}})^{2}=\dfrac{100}{9}\\\\\\a^{1}+2a^{\frac{1}{2}-\frac{1}{2}}+a^{-1}=\dfrac{100}{9}\\\\\\a+2a^{0}+a^{-1}=\dfrac{100}{9}\\\\\\a+a^{-1}+2=\dfrac{100}{9}\\\\\\a+a^{-1}=\dfrac{100}{9}-2\\\\\\a+a^{-1}=\dfrac{100-18}{9}\\\\\\\boxed{\boxed{a+a^{-1}=\dfrac{82}{9}}}$