Question

19- Sejam r1 e r2 as raízes da equação de 2º grau 2x² - 6x + 3 = 0. Determine o valor de:
a) r1 + r2
b) r1 . r2
c) (r1 + 3).(r2 + 3)
d) 1/r1 + 1/r2
e) r1² + r2²

Answer 1

$2x^2-6x+3=0$
Δ = $(-6)^2-4*2*3$
Δ = $36 - 24$
Δ = $12$

$r1= \frac{-(-6)+ \sqrt{12} }{2*2}= \frac{6+2 \sqrt{3} }{4}= \frac{3\ +\ \sqrt{3}}{2}$

$r2 = \frac{-(-6)- \sqrt{12} }{4}= \frac{6-2* \sqrt{3} }{4} = \frac{3\ -\ \sqrt{3} }{2}$

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

$b) \frac{3+ \sqrt{3} }{2}* \frac{3\ -\ \sqrt{3} }{2}= \frac{9-3 \sqrt{3}+3 \sqrt{3}-3 }{4}= \frac{9-3}{4}= \frac{6}{4}= \frac{3}{2}$

$c) ( \frac{3+ \sqrt{3} }{2}+3)*( \frac{3- \sqrt{3} }{2}+3) = \frac{9+ \sqrt{3} }{2}* \frac{9- \sqrt{3} }{2}= \frac{81-9 \sqrt{3}+9 \sqrt{3}-3 }{4}= \frac{81-3}{4}=$$\frac{78}{4}= \frac{39}{2}$

$d) \frac{2}{3+ \sqrt{3} }+ \frac{2}{3- \sqrt{3} }= \frac{6-2 \sqrt{3}+6+2 \sqrt{3} }{(3+ \sqrt{3})*(3- \sqrt{3}) } = \frac{12}{9-3 \sqrt{3}+3 \sqrt{3}-3 }= \frac{12}{6}=2$

$e) ( \frac{3+ \sqrt{3} }{2})^2+( \frac{3- \sqrt{3} }{2})^2=( \frac{12+6 \sqrt{3}}{4})+( \frac{12-6 \sqrt{3} }{4})= \frac{48+24 \sqrt{3}+48-24 \sqrt{3} }{16} = \frac{96}{16}$$=6$