Question

3 x elevado a 2 - x - 1 = 0

Answer 1

Resposta:

OLÁ

VAMOS A SUA PERGUNTA:⇒⇒

$\sf \: 3x^{2}-x-1=0$

$\sf \: 3x^{2}-x-1-\left(-1\right)=-\left(-1\right)$

$\sf \: 3x^{2}-x=-\left(-1\right)$

$\sf \: 3x^{2}-x=1$

$\sf \: \dfrac{3x^{2}-x}{3}=\dfrac{1}{3}$

$\sf \: x^{2}+\dfrac{-1}{3}x=\dfrac{1}{3}$

$\sf \: x^{2}-\dfrac{1}{3}x=\dfrac{1}{3}$

$\sf \: x^{2}-\dfrac{1}{3}x+\left(-\dfrac{1}{6}\right)^{2}=\dfrac{1}{3}+\left(-\dfrac{1}{6}\right)^{2}$

$\sf \: x^{2}-\dfrac{1}{3}x+\dfrac{1}{36}=\dfrac{1}{3}+\dfrac{1}{36}$

$\sf \: x^{2}-\dfrac{1}{3}x+\dfrac{1}{36}=\dfrac{13}{36}$

$\sf \: \left(x-\dfrac{1}{6}\right)^{2}=\dfrac{13}{36}$

$\sf \: \sqrt{\left(x-\dfrac{1}{6}\right)^{2}}=\sqrt{\dfrac{13}{36}}$

$\sf \: x-\dfrac{1}{6}=\dfrac{\sqrt{13}}{6} x-\dfrac{1}{6}=-\dfrac{\sqrt{13}}{6}$

$\boxed{\bold{\displaystyle{\clubsuit\ \spadesuit\ \maltese \sf \red{ x=\frac{\sqrt{13}+1}{6}\approx 0.767591879} }}}\ \checkmark \ \ \\ \boxed{\bold{\displaystyle{\clubsuit\ \spadesuit\ \maltese\ \sf \green{x=\frac{1-\sqrt{13}}{6}\approx -0.434258546}  }}}\ \checkmark$

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