Question

Sejam r e s as raízes da equação de 2 grau 2x²-6x+3 =0 determine o valor de:
a) r+s
b)r.s
c)(r+3).(s+3)
d)1÷r+1÷s

Answer 1

a)
$a) \ r+s=\frac{-(-6)}{2}=3\\ \\ b) \ r.s=\frac{3}{2}\\ \\ c) (r+3)(s+3)=r s+3 r+3 s+9=rs+3(r+s)+9=\frac{3}{2}+3*3+9\\ \\=18+\frac{3}{2}=\frac{39}{2}\\ \\ d) \ \frac{1}{r}+\frac{1}{s}=\frac{r+s}{rs}=\frac{3}{\frac{3}{2}}=3*\frac{2}{3}=2$

Answer 2

$2x^2-6x+3=0$

$r+s=\dfrac{-b}{a}$

$r+s=\dfrac{-(-6)}{2}$

$r+s=\dfrac{6}{2}$

$r+s=3$

b) $rs=\dfrac{c}{a}$

$rs=\dfrac{3}{2}$

c) $(r+3)(s+3)=rs+3r+3s+9=rs+3(r+s)+9$

$(r+3)(s+3)=\dfrac{3}{2}+3\cdot3+9$

$(r+3)(s+3)=\dfrac{3}{2}+9+9$

$(r+3)(s+3)=\dfrac{3+18+18}{2}$

$(r+3)(s+3)=\dfrac{39}{2}$

d) $\dfrac{1}{r}+\dfrac{1}{s}=\dfrac{r+s}{rs}$

$\dfrac{1}{r}+\dfrac{1}{s}=\dfrac{3}{\dfrac{3}{2}}$

$\dfrac{1}{r}+\dfrac{1}{s}=\dfrac{3}{1}\cdot\dfrac{2}{3}$

$\dfrac{1}{r}+\dfrac{1}{s}=2$