Question

9x^2-12x+4=0
8^{x-2}=\sqrt{8}
7x^2+21x-28<0
ayuda con estos problemas

Answer 1

Respuesta:

$9x^2-12x+4=0\\\\x_{1,\:2}=\frac{-\left(-12\right)\pm \sqrt{\left(-12\right)^2-4\cdot \:9\cdot \:4}}{2\cdot \:9}\\\\x_{1,\:2}=\frac{-\left(-12\right)\pm \sqrt{0}}{2\cdot \:9}\\\\x=\frac{-\left(-12\right)}{2\cdot \:9}\\\\x=\frac{2}{3}$

$8^{x-2}=\sqrt{8}\\\\x-2=\frac{1}{2}\\\\x=\frac{5}{2}$

$7x^2+21x-28<0\\\\x^2+3x-4<0\\\\\left(x-1\right)\left(x+4\right)<0\\\\-4<x<1$

Answer 2

Respuesta:

das corona y te ayudo