Question

não sei resolver esses itens,me ajudemmm​

Answer 1

Resposta:

$Formula~de~Bhaskara\\ \\2x^{2}-6x+3=0\\ \\delta=b^{2}-4.a.c\\ \\delta=(-6)^{2}-4.2.3\\delta=36-24\\ \\delta=12\\ \\x=\frac{-b+-\sqrt{12}}{2.2}\\ \\x=\frac{6+-\sqrt{12} }{4}\\ \\x'=r_{1}=\frac{6+\sqrt{12}}{4}\\ \\x''=r_{2}=\frac{6-\sqrt{12}}{4}$

$a)\\ \\ r_{1}+r_{2}=\frac{6+\sqrt{12}}{4}+\frac{6-\sqrt{12}}{4}=\frac{6+\sqrt{12}+6-\sqrt{12}}{4}=\frac{12}{4}=3$

$b)\\ \\r_{1}.r_{2}=(\frac{6+\sqrt{12}}{4}).(\frac{6-\sqrt{12}}{4})=\frac{36-6\sqrt{12}+6\sqrt{12}-(\sqrt{12})^{2}}{16}=\\ \\\frac{36+(\sqrt{12})^{2}}{16}=\frac{36+12}{16}=\frac{48}{16}=3$

$c)\\ \\(r_{1}+3).(r_{2}+3)=(\frac{6+\sqrt{12}}{4}+3).(\frac{6-\sqrt{12}}{4}+3)=\\ \\(\frac{6+\sqrt{12}}{4}).(\frac{6-\sqrt{12}}{4})+3(\frac{6+\sqrt{12}}{4})+3.(\frac{6-\sqrt{12}}{4})=\\ \\\frac{6^{2}-(\sqrt{12})^{2}}{16}+18+\frac{3\sqrt{12}}{4}+18-\frac{3\sqrt{12}}{4}=\\ \\\frac{36-12}{16}+18+18=\frac{24}{16}+36=\frac{24+576}{16}=\frac{600}{16}=37,5$

$d)\\ \\\frac{1}{r_{1}}+\frac{1}{r_{2}}=\frac{1}{\frac{6+\sqrt{12}}{4}}+\frac{1}{\frac{6-\sqrt{12}}{4}}=\frac{4}{6+\sqrt{12}}+\frac{4}{6-\sqrt{12}}\\ \\mmc=(6+\sqrt{12})(6-\sqrt{12})\\ \\\frac{4(6-\sqrt{12})+4(6+\sqrt{12})}{(6+\sqrt{12})(6-\sqrt{12})}=\frac{24-4\sqrt{12}+24+4\sqrt{12}}{6^{2}-(\sqrt{12})^{2}}=\frac{48}{36-12}=\frac{48}{24}=2$

$e)\\ \\(r_{1})^{2}+(r_{2})^{2}=(\frac{6+\sqrt{12}}{4})+(\frac{6-\sqrt{12}}{4})=\\ \\ \frac{6+\sqrt{12}+6-\sqrt{12}}{4}=\frac{12}{4}=3$

Estude, aprenda e aplique todas as regras de produtos notáveis. Assim você não terá dificuldades em grande parte da matemática.

Espero ter ajudado.

Bom fim de semana!

desculpe a e) é ao quadrado

$e)~~~~corrigindo\\  \\ \\(6+\frac{\sqrt{12}}{4})^{2}+(6-\frac{\sqrt{12}}{4})^{2}=\\ \\6^{2}+\frac{12}{16}+6^{2}-\frac{12}{16}=36+36=72$

Agora sim!

Bons estudos e Bom fim de semana!