Question

calcular os zeros das seguintes funções:
a) f(x)=-3x-10
b)f(x)=x²+x-20
c)f(x)=-x²-x+12

Answer 1

Vamos lá Elias,

(A)

f(x) =x² - 3x - 10
x² - 3x - 10 = 0
Δ = (-3)² - 4 * 1 * (-10)
Δ = 9 + 40
Δ = 49

$x=\frac{-(-3)\pm\sqrt{49}}{2\cdot 1}\to x=\frac{3\pm7}{2}\\ \\ x'=\frac{10}{2}\to x'=5\\ \\ x''=\frac{-4}{2}\to x''=-2\\ \\ \boxed{\boxed{S=\{5,-2\}}}$

(B)

f(x) = x² + x - 20
x² + x - 20 = 0
Δ = 1² - 4 * 1 * (-20)
Δ = 1 + 80
Δ = 81

$x=\frac{-1\pm\sqrt{81}}{2\cdot 1}\to x=\frac{-1\pm 9}{2}\\ \\ x'=\frac{8}{2}\to x'=4\\ \\ x''=\frac{-10}{2}\to x=-5\\ \\ \boxed{\boxed{S=\{4,-5\}}}$

(C)

f(x) = -x² - x + 12
-x² - x + 12 = 0
Δ = (-1)² - 4 * (-1) * 12
Δ = 1 + 48
Δ = 49

$x=\frac{-(-1)\pm\sqrt{49}}{2\cdot (-1)}\to x=\frac{1\pm 7}{-2}\\ \\ x'=\frac{8}{-2}\to x'=-4\\ \\ x''=\frac{-6}{-2}\to x''=3\\ \\ \boxed{\boxed{S=\{-4,3\}}}$

Espero ter ajudado (:
Bons estudos!