Question

qual é o valor de. x2+9x+8=0

Answer 1

$x^{2} +9x+8=0$

Δ = $9^{2} -4*8$

Δ= $9^{2} -32$

Δ = $81-32$

Δ = $49$

$x_{1,2} = \frac{(-9)\± \sqrt{49} }{2}$

$x_{1,2} = \frac{(-9)\±7}{2}$

$x' = \frac{(-9)+7}{2}$

$x'= \frac{-2}{2}$

$x'=-1$

$x''= \frac{-9-7}{2}$

$x''= \frac{-16}{2}$

$x''=-8$

Answer 2

x² + 9x + 8 = 0

a = 1; b = 9; c = 8

Delta:
Δ = b² - 4ac
Δ = 9² - 4 * 1 * 8
Δ = 81 - 32
Δ = 49
       Bhaskara:
       x = - b ± √Δ / 2a
       x = - 9 ± √49 / 2 * 1
       x = - 9 ± 7 / 2
       x' = - 9 + 7 / 2 = -2 / 2 = -1
       x'' = - 9 - 7 / 2 = -16 / 2 = ⇒-8
S = {-8, -1}

Espero ter ajudado. Valeu!