Question

C) 4 + x (x - 4) = 3x​

Answer 1

Resposta:

Explicação passo a passo:

C) 4 + x.(x - 4) = 3x​

4 + x^2 - 4x = 3x

x^2 - 4x - 3x + 4 = 0

x^2 - 7x + 4 = 0

a = 1; b = - 7; c = 4

/\= b^2 - 4ac

/\= (-7)^2 - 4.1.4

/\ = 49 - 16

/\= 33

X = [- b +/- \/ /\]/ 2a

X = [ -(-7) +/- \/33] / 2.1

X' = (7 + \/33)/2

X " = (7 - \/33)/2

Answer 2

Resposta:

$x_{1} = \frac{7 + \sqrt{33} }{2} \: \: \: \: \: \: \: x_{2} = \frac{7 - \sqrt{ 33} }{2}$

Explicação passo-a-passo:

$\blue{4 + x(x - 4) = 3x} \\4 + x(x - 4) = 3x \\4 + x(x - 4) \\4 + x {}^{2} - 4x \\x {}^{2} - 4x + 4 \\4 + x(x - 4) = 3x \\x {}^{2} - 4x + 4 = 3x \\x {}^{2} - 4x + 4 - 3x = 3x - 3x \\x {}^{2} - 4x + 4 - 3x = 0 \\x {}^{2} - 4x + 4 = 3x\\x {}^{2} - 4x + 4 - 3x = 0 \\ - 4x - 3x \\( - 4 - 3)x \\ - 7x \\x {}^{2} - 4x + 4 - 3x = 0\\x {}^{2} - 7x + 4 = 0 \\x {}^{2} - 2 \times \frac{7}{2} \times + 4x \\ ( \frac{7}{2}) {}^{2} - ( \frac{7}{2}) {}^{2} = 0 \\(x - \frac{7}{2}) {}^{2} + 4 - ( \frac{7}{2}) {}^{2} = 0 \\x {}^{2} - 7x + 4 = 0 \\ (x - \frac{7}{2}) {}^{2} + 4 - ( \frac{7}{2}) {}^{2} = 0 \\ (x - \frac{7}{2}) {}^{2} = - 4 + ( \frac{7}{2}) {}^{2} \\ (x - \frac{7}{2}) = - 4 + \frac{7 {}^{2} }{2 {}^{2} } \\ (x - \frac{7}{2}) {}^{2} = - 4 + \frac{49}{2 {}^{2}} \\ (x - \frac{7}{2}) {}^{2} = - 4 + \frac{49}{4} \\ (x - \frac{7}{2}) {}^{2} = \frac{33}{4} \\ (x - \frac{7}{2}) {}^{2} - 4 + \frac{7 {}^{2} }{2 {}^{2} } \\ (x - \frac{7}{2}) {}^{2} = \frac{33}{4} \\ x = - \frac{7}{2} = + \sqrt{ \frac{33}{2} } \\ x - \frac{7}{2} = + \sqrt{ \frac{33}{2} } \\ x = + \sqrt{ \frac{33}{2} } + \frac{7}{2} \\ x = - \frac{7}{2} = + \frac{ \sqrt{33} }{2} \\ x = + \frac{ \sqrt{33} }{2} + \frac{7}{2} \\ x_{} = \frac{7}{2} = + \frac{ \sqrt{33} }{2} \\x = \frac{7}{2} - \sqrt{ \frac{33}{2} } \\ x = 7 + \sqrt{ \frac{33}{2} } \\ x = \frac{7}{2} - \frac{ \sqrt{33} }{2} \\ x = 7 - \frac{ \sqrt{33} }{2} \\ x = \frac{7}{2} + \frac{ \sqrt{33} }{2} \\ x = 7 - \sqrt{ \frac{33}{2} } \\ \green{resposta} \\ \purple{x _{1} = \frac{7 + \sqrt{33} }{2} } \\ \purple{x_{2} = \frac{7 - \sqrt{33} }{2} }$