Question

a soma dos inversos das raizes da equaçao x2-2x-1=0 sao?

Answer 1

Resposta:

Explicação passo-a-passo:

$x=\frac{-b \pm \sqrt{b^2-4ac} }{2a}= \frac{-(-2) \pm \sqrt{(-2)^2-4\cdot 1 \cdot (-1)} }{2 \cdot 1} =\frac{2\pm \sqrt{4+4} }{2}= \frac{2\pm \sqrt{8} }{2}= \frac{2\pm 2\sqrt{2}}{2}= 1\pm \sqrt{2}\\\\x'=1+ \sqrt{2}\\\\x''= 1- \sqrt{2}\\\\Soma \ dos\ inversos: \ \frac{1}{x'}+\frac{1}{x''} \\\\=\frac{1}{ 1+ \sqrt{2}}+\frac{1}{ 1- \sqrt{2}} =\frac{1- \sqrt{2}}{\big(1+ \sqrt{2}\big)\cdot \big(1- \sqrt{2}\big)} +\frac{1+ \sqrt{2}}{\big(1+ \sqrt{2}\big)\cdot \big(1- \sqrt{2}\big)}$

$=\frac{1- \sqrt{2}}{1- \sqrt{4}} +\frac{1+ \sqrt{2}}{1- \sqrt{4}}=\frac{1- \sqrt{2}}{1- 2} +\frac{1+ \sqrt{2}}{1- 2}=\frac{1- \sqrt{2}}{-1} +\frac{1+ \sqrt{2}}{-1}\\\\\Big(-1+ \sqrt{2} \Big)+\Big(-1- \sqrt{2}\Big)=-1+ \sqrt{2}-1- \sqrt{2}=-2$