Question

2) - 1) Determine as raízes das equações a seguir:
a) x2 - 5x + 6 = 0
b) 2y2 + 3y - 14 = 0



me ajudeeem​

Answer 1

Explicação passo-a-passo:

Determinar as raízes (valor de x ou y) das equações:

a)

$\sf{x^2 - 5x + 6 = 0}$

$\sf{a = 1~~b = - 5~~ c = 6}$

$\sf{\Delta = b^2 - 4ac}$

$\sf{\Delta = (-5)^2 - 4\cdot1\cdot6}$

$\sf{\Delta = 25 - 24}$

$\sf{\Delta = 1}$

$\sf{x = \dfrac{- b \pm \sqrt{\Delta}}{2a}}$

$\sf{x = \dfrac{- (-5) \pm \sqrt{1}}{2\cdot1}}$

$\sf{x = \dfrac{5 \pm 1}{2}}$

$•~~\sf{x' = \dfrac{5 + 1}{2} = \dfrac{6}{2} = 3}$

$•~~\sf{x" = \dfrac{5 - 1}{2} = \dfrac{4}{2} = 2}$

$\boxed{\sf{S = \left\{2~~;~~3\right\}}}$

As raízes são 3 e 2

b)

$\sf{2y^2 + 3y - 14 = 0}$

$\sf{a = 2~~b = 3~~ c = - 14}$

$\sf{\Delta = b^2 - 4ac}$

$\sf{\Delta = 3^2 - 4\cdot2\cdot(-14)}$

$\sf{\Delta = 9 + 112}$

$\sf{\Delta = 121}$

$\sf{y = \dfrac{- b \pm \sqrt{\Delta}}{2a}}$

$\sf{y = \dfrac{- 3 \pm \sqrt{121}}{2\cdot2}}$

$\sf{y = \dfrac{- 3 \pm 11}{4}}$

$•~~\sf{y' = \dfrac{- 3 + 11}{4} = \dfrac{8}{4} = 2}$

$•~~\sf{y" = \dfrac{- 3 - 11}{4} = - \dfrac{14}{4} = - \dfrac{14\div2}{4\div2} = - \dfrac{7}{2}}$

$\boxed{\sf{S = \left\{- \dfrac{7}{2}~~;~~2\right\}}}$

As raízes são 2 e - 7/2