Question

considere os números 3;2;1;e 0.5.dois deles são a soluçoes de 2x²-5+2:0 .quais são

Answer 1

$2x^2 - 5x + 2 = 0$

Resolvendo por Bháskara

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=2, b=−5, c=2

Δ=b²−4ac
Δ=(−5)²−4*(2)*(2)
Δ=25−16
Δ=9

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a}$

$x = \dfrac{-(-5) \pm \sqrt{9}}{2*2} \\ \\ \\ x = \dfrac{5 \pm 3}{4} \\ \\ \\ x' = \dfrac{5 + 3}{4} \\ \\ x' = \dfrac{8}{4} \\ \\ \\ x' = 2 \\ \\ \\ x'' = \dfrac{5 - 3}{4} \\ \\ \\ x'' = \dfrac{2}{4} \\ \\ \\ x'' = \dfrac{1}{2}$

S = {$2, \ \dfrac{1}{2}$}

====

$\dfrac{1}{2} = 0,5$


Resposta  = 2 e 0,5