Question

Determine o valor de x,sabendo que x= $\frac{-b+ \sqrt{ b^{2-4ac} } }{2a}$ , quando:
a) a =1 b=5 e c=6
a) a =1 b=-5 e c=-6
a) a =2 b=3 e c=-2
a) a =-1 b=-7 e c=-10
a) a =1 b=8 e c=16
a) a =1 b=2 e c=15
a) a =-3 b=1 e c=2

Answer 1

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

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

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x² -5x + 6 = 0

Δ=b²−4ac
Δ=(5)²−4⋅1⋅6
Δ=25−24
Δ=1

$x = \dfrac{-b \pm \sqrt{\Delta}}{2*a}\\ \\ \\ x = \dfrac{-5 \pm \sqrt{1}}{2*1}\\ \\ \\ x = \dfrac{-5 \pm 1}{2}\\ \\ \\ x' = \dfrac{-5 - 1}{2}\\ \\ \\ x' = \dfrac{-6}{2}\\ \\ \\ x' = -3 \\ \\ \\ x'' = \dfrac{-5 + 1}{2}\\ \\ \\ x'' = \dfrac{-4}{2}\\ \\ \\ x'' = -2$

S = {-3,  -2}

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x² -5x -6 = 0

Δ = b²−4ac
Δ = (−5)²−4⋅1⋅−6
Δ = 25+24
Δ = 49

$x = \dfrac{-b \pm \sqrt{\Delta}}{2*a}\\ \\ \\x = \dfrac{-(-5) \pm \sqrt{49}}{2*1}\\ \\ \\x = \dfrac{5 \pm 7}{2}\\ \\ \\x' = \dfrac{5 - 7}{2}\\ \\ \\x' = \dfrac{-2}{2}\\ \\ \\x' = -1 \\ \\ \\ x'' = \dfrac{5 + 7}{2}\\ \\ \\x'' = \dfrac{12}{2}\\ \\ \\x'' = 6$

S = {-1, 6}

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2x² +3x - 2 = 

Δ = b²−4ac
Δ = (3)²−4⋅2⋅−2
Δ = 9+16
Δ = 25


$x = \dfrac{-b \pm \sqrt{\Delta}}{2*a}\\ \\ \\x = \dfrac{-3 \pm \sqrt{25}}{2*2}\\ \\ \\x = \dfrac{-3 \pm 5}{4}\\ \\ \\x' = \dfrac{-3 - 5}{4}\\ \\ \\x' = \dfrac{-8}{4}\\ \\ \\x' = -2 \\ \\ \\ x'' = \dfrac{-3 + 5}{2}\\ \\ \\x'' = \dfrac{2}{4}\\ \\ \\x'' = \dfrac{1}{2}$

S = {$-2, ~ \dfrac{1}{2}$}

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-x² - 7x - 10 = 0

Multiplicar por -1 para retirar o sinal de negativo do primeiro termo:Nõ altera o resultado:

x² + 7x + 10 = 0

Δ = b²−4ac
Δ = (7)2−4⋅1⋅10
Δ = 49−40
Δ = 9

$x = \dfrac{-b \pm \sqrt{\Delta}}{2*a}\\ \\ \\x = \dfrac{-7 \pm \sqrt{9}}{2*1}\\ \\ \\x = \dfrac{-7 \pm 3}{2}\\ \\ \\x' = \dfrac{-7 - 3}{2}\\ \\ \\x' = \dfrac{-10}{2}\\ \\ \\x' = -5 \\ \\ \\x'' = \dfrac{-7 + 3}{2}\\ \\ \\x'' = \dfrac{-4}{2}\\ \\ \\x'' = -2$

S = {-5,  -2}

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x² + 8x + 16 = 0

Δ = b²−4ac
Δ = (8)²−4⋅1⋅16
Δ = 64−64
Δ = 0

$x = \dfrac{-b \pm \sqrt{\Delta}}{2*a}\\ \\ \\x = \dfrac{-8 \pm \sqrt{0}}{2*1}\\ \\ \\x = \dfrac{-8 \pm 0}{2}\\ \\ \\x' = \dfrac{-8 - 0}{2}\\ \\ \\x' = \dfrac{-8}{2}\\ \\ \\x' = -4 \\ \\ \\ x'' = \dfrac{-8 + 0}{2}\\ \\ \\x'' = \dfrac{-8}{2}\\ \\ \\x'' = -4$

S = {-4,  -4}

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x² + 2x + 15 = 0 

Δ = b²−4ac
Δ = (2)²−4⋅1⋅15
Δ = 4−60
Δ = −56

Sem solução para o conjuntos dos números reais

Resolvendo para o conjunto dos números complexos:


$x = \dfrac{-b \pm \sqrt{\Delta}}{2*a}\\ \\ \\x = \dfrac{-2 \pm \sqrt{-56}}{2*1}\\ \\ \\x = \dfrac{-2 \pm 2\sqrt{14}. i}{2}\\ \\ \\x' = \dfrac{-2 - 2\sqrt{14}. i}{2} \\ \\ \\ x' = -1-\sqrt{14}. i \\ \\ \\ x'' = \dfrac{-2 - 2\sqrt{14}. i}{2} \\ \\ \\ x'' = -1+\sqrt{14}. i$

S = { $~ -1-\sqrt{14}. i, ~~ -1+\sqrt{14}. i ~$ }

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-3x² + x + 2 = 0

Multiplicar por -1

3x² -x - 2 = 0

Δ = b²−4ac
Δ = (−1)²−4⋅3⋅−2
Δ = 1+24
Δ = 25

$x = \dfrac{-b \pm \sqrt{\Delta}}{2*a}\\ \\ \\x = \dfrac{-(-1) \pm \sqrt{25}}{2*3}\\ \\ \\x = \dfrac{1 \pm 5}{6}\\ \\ \\x' = \dfrac{1 - 5}{6}\\ \\ \\x' = \dfrac{-4}{6}\\ \\ \\x' = \dfrac{-2}{3}\\ \\ \\x'' = \dfrac{1 + 5}{6}\\ \\ \\x'' = \dfrac{6}{6}\\ \\ \\x'' = 1$

S = {$\dfrac{-2}{3}. ~~1$}