Question

Resuelve la siguiente operación completando el Trinomio Cuadrado Perfecto: 6x^2-x-12=0

Answer 1

¡Hola!

$6 {x}^{2} - x - 12 = 0 \\ \\ \dfrac{6{x}^{2}}{6} -\dfrac{1x}{6} -\dfrac{12}{6} =\dfrac{0}{0} \\\\ {x}^{2}-\dfrac{1}{6}x-2=0 \\\\{x}^{2}-\dfrac{1}{6}x=2$

Buscamos la mitad de 1/6.

$\dfrac{\frac{1}{6}}{\frac{2}{1}}= \dfrac{1\times1}{6\times12}=\big(\dfrac{1}{12} \big)^2 \\\\\\ {x}^{2}+\big(\dfrac{1}{12} \big)^2-\dfrac{1}{6}x=2+\big(\dfrac{1}{12} \big)^2 \\\\ \big(x+\dfrac{1}{12} \big)^2=2+\dfrac{1}{144} \\\\ \big(x+\dfrac{1}{12} \big)^2= \dfrac{2+1}{144} \\\\ \big(x+\dfrac{1}{12} \big)^2= \dfrac{3}{144} \\\\ \sqrt{\big(x+\dfrac{1}{12} \big)^2}= \sqrt{\dfrac{1}{48}} \\\\ x+\dfrac{1}{12}= \pm \dfrac{\sqrt{3} }{12}\\\\ x=\pm \dfrac{\sqrt{3}}{12}-\dfrac{1}{12}$

Resolver para x₁.

$x=+ \dfrac{\sqrt{3}}{12}-\dfrac{1}{12} \\\\ \boxed{x=\dfrac{\sqrt{3}-1}{12}}$

Resolver para x₂.

$x=- \dfrac{\sqrt{3}}{12}-\dfrac{1}{12} \\\\ \boxed{x=\dfrac{-\sqrt{3}-1}{12}}$

Concluimos que...

$\boxed{ x= \big\{ \frac{ \sqrt{3}-1}{12}, - \frac{ \sqrt{3}-1}{12} \big\} }$

Espero que sirva y saludos.