Question

Considere a equação x² –5x – 6 = 0 e sejam x' e x'' suas raízes. Então, (x')²+ (x'')² vale:

Answer 1

x² –5x – 6 = 0

delta = b² - 4.a.c

delta = (-5)² - 4.1.(-6)

delta = 25 + 24 = 1

x = ( - b +- raiz de delta ) / 2.a

x = ( - (-5) +- raiz de 1 ) / 2.1

x = ( 5 +- 1 ) / 2

x' = ( 5 + 1 ) / 2 = 6 / 2 = 3

x" = ( 5 - 1 ) / 2 = 4 / 2 = 2

(x')² + (x")² = (3)² + (2)² = 9 + 4 = 13

Answer 2

$\mathsf{x^2-5x+6=0}\\\mathsf{a=1~~b=-5~~c=6}$

$\mathsf{x'+x''=-\dfrac{b}{a}=-\dfrac{-5}{1}=5}\\\mathsf{x'.x''=\dfrac{c}{a}=\dfrac{6}{1}=6}$

$\mathsf{(x'+x'')^2=(x')^2+2.x'.x''+(x'')^2}\\\mathsf{(x')^2+(x')^2=(x'+x'')^2-2.x'.x''}$

$\mathsf{(x')^2+(x'')^2=5^2-2.(6)}\\\mathsf{(x')^2+(x'')^2=25-12}\\\large\boxed{\boxed{\boxed{\boxed{\mathsf{(x')^2+(x'')^2=13}}}}}$