Question

Determine as raízes das equações utilizando a fórmula resolutiva

Answer 1

Resposta:

Explicação passo-a-passo:

a)

$2x^{2} +x-1=0\\\\delta=b^{2} -4ac\\delta=1^{2} -4.2.(-1)\\delta=1+8\\delta=9\\\\x'=\frac{-b+\sqrt{delta} }{2a} \\\\x'=\frac{-1+3}{4} \\\\x'=\frac{2}{4} =\frac{1}{2} \\\\\\x"=\frac{-b-\sqrt{delta} }{2a} \\\\x"=\frac{-1-3}{4} \\\\x"=\frac{-4}{4} =-1$

S = { -1, 1/2 }

b)

$2x^{2} +2x-24=0\\\\delta=b^{2} -4ac\\delta=2^{2} -4.2.(-24)\\delta=4+192\\delta=196\\\\x'=\frac{-b+\sqrt{delta} }{2a} \\\\x'=\frac{-2+14}{4} \\\\x'=\frac{12}{4} =3\\\\\\x"=\frac{-b-\sqrt{delta} }{2a} \\\\x"=\frac{-2-14}{4} \\\\x"=\frac{-16}{4} =-4$

S = { -4, 3 }

c)

$3x^{2} -4x-2=-3\\3x^{2} -4x+1=0\\\\delta=b^{2} -4ac\\delta=(-4)^{2} -4.3.1\\delta=16-12\\delta=4\\\\x'=\frac{-b+\sqrt{delta} }{2a} \\\\x'=\frac{4+2}{6} \\\\x'=\frac{6}{6} =1\\\\\\x"=\frac{-b-\sqrt{delta} }{2a} \\\\x"=\frac{4-2}{6} \\\\x"=\frac{2}{6} =\frac{1}{3}$

S = { 1/3, 1 }

Espero ter ajudado!

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