Question

Si la ecuación 2nx²-(x+4)x+2=0 ,tiene raíces iguales, calcule n. XF URGENTE¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡

Answer 1

Respuesta:

$2nx2(x+4)x+2=0: x=\frac{4+\sqrt{-16n+24} }{2(2n-1)} , x=\frac{4-\sqrt{-16n+24} }{2(2n-1)}; n\ne \frac{1}{2}$

Explicación paso a paso:

$2nx^2-x^2-4x+2=0$

$\left(2n-1\right)x^2-4x+2=0$

$\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4\left(2n-1\right)\cdot \:2}}{2\left(2n-1\right)}: \frac{4+\sqrt{-16n+24} }{2(2n-1)}$

$\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4\left(2n-1\right)\cdot \:2}}{2\left(2n-1\right)}: \frac{4-\sqrt{-16n+24} }{2(2n-1)}$

$x= \frac{4+\sqrt{-16n+24}}{2\left(2n-1\right)},\:x=\frac{4-\sqrt{-16n+24}}{2\left(2n-1\right)};\quad \:n\ne \frac{1}{2}$

(Espero que te haya servido de ayuda)