Matemática, perguntado por LuuhStyles, 1 ano atrás

Determine as raízes ( ou zeros) das funções:
a) f(x)= x² + 2x - 3
b) f(x)= -4x² + 4x - 1

Soluções para a tarefa

Respondido por danielfalves
1
a)

f(x)=x^2+2x-3\\\\x^2+2x-3=0\\\\a=1\,\,\,\,\,\,\,\,\,\,b=2\,\,\,\,\,\,\,\,\,\,c=-3\\\\\triangle=b^2-4ac\\\triangle=(2)^2-4.(1).(-3)\\\triangle=4+12\\\triangle=16\\\\x= \dfrac{-b \frac{+}{-} \sqrt{\triangle}  }{2a}\\\\\\x= \dfrac{-2 \frac{+}{-} \sqrt{16}  }{2}\\\\\\x= \dfrac{-2 \frac{+}{-}4 }{2}\\\\\\x'= \dfrac{2}{2}\\\\\\\boxed{x'=1}\\\\x"= -\dfrac{6}{2} \\\\\boxed{x"=-3}

b)

f(x)=-4x^2+4x-1\\\\-4x^2+4x-1=0\\\\a=-4\,\,\,\,\,\,\,\,\,\,b=4\,\,\,\,\,\,\,\,\,\,c=-1\\\\\triangle=b^2-4ac\\\triangle=(4)^2-4.(-4).(-1)\\\triangle=16-16\\\triangle=0\\\\x= \dfrac{-b \frac{+}{-} \sqrt{\triangle}  }{2a}\\\\\\x=  \dfrac{-4 \frac{+}{-} \sqrt{0}  }{2.(-4)}\\\\\\x= \dfrac{-4}{-8}\\\\\\\boxed{x'=x"= \dfrac{1}{2}}

LuuhStyles: nao entendi nd do q vc escreveu
danielfalves: Você está pelo aplicativo?
danielfalves: acesse pelo navegador... http://brainly.com.br/tarefa/7225064
LuuhStyles: ook vlw
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