Question

Me pide con una sola variable, ayuda!!!!! ​

Answer 1

¡Hola!

Respuesta:

$06: \ x = \frac{3}{4}, \ x = -4\\\\07: \ x = 4, \ x = -4\\\\08: \ x = \frac{3}{2}, \ x =- \frac{1}{2}$

Explicación paso a paso:

06.

$4x^2\:+\:13x\:-\:12\:=\:0\\\\x_{1,\:2}=\frac{-13\pm \sqrt{13^2-4\cdot \:4\left(-12\right)}}{2\cdot \:4}\\\\x_{1,\:2}=\frac{-13\pm \:19}{2\cdot \:4}\\\\Separar\:las\:soluciones:\\\\x_1=\frac{-13+19}{2\cdot \:4},\:x_2=\frac{-13-19}{2\cdot \:4}\\\\x=\frac{3}{4},\:x=-4$

07.

$\frac{x^2\:+\:2}{3}-\frac{x^2}{2}+2\:=0\\\\\frac{x^2+2}{3}-\frac{x^2}{2}=-2\\\\2\left(x^2+2\right)-3x^2=-12\\\\-x^2+4=-12\\\\-x^2=-16\\\\\mathrm{Dividir\:ambos\:lados\:entre\:}-1:\\\\\frac{-x^2}{-1}=\frac{-16}{-1}\\\\x^2=16\\\\\mathrm{Para\:}x^2=f\left(a\right)\mathrm{\:las\:soluciones\:son\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}:\\\\x=\sqrt{16},\:x=-\sqrt{16}\\\\x=4,\:x=-4$

08.

$\left(2x+3\right)\left(4x-5\right)+\left(8x-1\right)\left(2x-3\right)=6\\\\24x^2-24x-12=6\\\\24x^2-24x-18=0\\\\x_{1,\:2}=\frac{-\left(-24\right)\pm \sqrt{\left(-24\right)^2-4\cdot \:24\left(-18\right)}}{2\cdot \:24}\\\\x_{1,\:2}=\frac{-\left(-24\right)\pm \:48}{2\cdot \:24}\\\\\mathrm{Separar\:las\:soluciones:}\\\\x_1=\frac{-\left(-24\right)+48}{2\cdot \:24},\:x_2=\frac{-\left(-24\right)-48}{2\cdot \:24}\\\\x=\frac{3}{2},\:x=-\frac{1}{2}$