Question

2x²+20x+48=O
x²-8x+15=0

Answer 1

a)

$2x^2 + 20x + 48 = 0$

Para facilitar os cálculos podemos dividir a equação por 2, não altera o resultado:

$x^2 + 10x + 24 = 0$

Δ=b²−4acΔ=10²−4⋅1⋅24Δ=100−96Δ=4

$x = \dfrac{-b \pm \sqrt{\Delta}}{2*a} \\ \\ x = \dfrac{-10 \pm \sqrt{4}}{2*1}\\ \\ x = \dfrac{-10 \pm 2}{2}\\ \\x' = \dfrac{-10 + 2}{2}\\ \\ x' = \dfrac{-8}{2}\\ \\ x' = -4\\ \\ \\ x'' = \dfrac{-10 - 2}{2}\\ \\ x'' = \dfrac{-12}{2}\\ \\ x'' = -6$

S = {-4,  -6}

===
b)

$x^2 - 8x + 15 = 0$

Δ=b²2−4acΔ=(−8)²−4⋅1⋅15Δ=64−60Δ=4

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


$x = \dfrac{-(-8) \pm \sqrt{4}}{2*1}\\ \\ x = \dfrac{8 \pm 2}{2}\\ \\x' = \dfrac{8 + 2}{2}\\ \\ x' = \dfrac{10}{2}\\ \\ \\ \\ x' = 5\\ \\ x'' = \dfrac{8 - 2}{2}\\ \\ x'' = \dfrac{6}{2}\\ \\ x'' = 3$

S = {5, 3}