Question

como se resolver isso aqui?
$(x - 1) {}^{2} - x(x + 1) \div 3 = \frac{1}{2}$
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Answer 1

Vamos lá

$(x-1)^2 - \frac{x(x+1)}{3}  = \frac{1}{2}\\\frac{3(x-1)^2-x^2-x}{3} = \frac{1}{2}\\\frac{3(x^2-2x+1)-x^2-x}{3} =\frac{1}{2}\\\frac{3x^2-6x+3-x^2-x}{3} = \frac{1}{2}\\\frac{2x^2-7x+3}{3}=\frac{1}{2}\\2x^2-7x+3 = 1,5\\2x^2-7x+1,5=0\\4x^2-14x+3=0$

Aplicando Bhaskara:

$\triangle=(-14)^2-4*4*3 = 196-48=148\\\\\frac{-(-14)+-\sqrt{148}}{2*4} = \frac{14+-2\sqrt{37}}{8}$

$x' = \frac{7+\sqrt{37}}{4} \\x'' = \frac{7-\sqrt{37}}{4}$