Question

Considere x1 e x2 as raizes da equação : 3x2 - 7x + 15=0. Calcule

A- x1 + x2

B- x1 . X2

C- 1/x1 + 1/x2

D- x elevado a 2 e 1 + x elevado a 2 e 2

E- x elevado a 3 e 1 + x elevado a 3 e 2

Answer 1

3x²-7x+15=0

x1+x2=-b/a=-(-7)/3=7/3

x1.x2=c/a=15/3=5

a)

x1+x2=7/3

b) x1.x2=5

c)

$\frac{1}{x1} + \frac{1}{x2} = \frac{x1 + x2}{x1.x2} = \frac{ \frac{7}{3} }{5} = \frac{7}{3}. \frac{1}{5} \\ = \frac{7}{15}$

d)

${x1}^{2} + {x2}^{2} \\ = {(x1 + x2 )}^{2} - 2.x1.x2 \\ = {( \frac{7}{3} )}^{2} - 2.5 = \frac{49}{9} - 10$

$\frac{49}{9} - 10 = \frac{49 - 90}{9} = - \frac{41}{9}$

e)

(x1+x2)³=x1³+3.x1².x2+3.x1.x2²+x2³

(x1+x2)³=x1³+x2³+ 3.x1.x2(x1+x2)

x1³+x2³=(x1+x2)³-3.x1.x2(x1+x2)

${x1}^{3} + {x2}^{3} = { (\frac{7}{3}) }^{3} - 2.5.( \frac{7}{3}) \\ {x1}^{3} + {x2}^{3} = \frac{343}{27} - \frac{70}{3}$

${x1}^{3} + {x2}^{3} = \frac{343 - 630}{27} = - \frac{287}{27}$