[tex]3b)\\\\ \frac{1}{4}x^2-3x+5=( \frac{1}{2}x-4)( \frac{1}{2}x +4)\\\\\frac{1}{4}x^2-3x+5=\frac{1}{4}x^2-16\\\\\frac{1}{4}x^2-3x -\frac{1}{4}x^2=-16 -5\\\\-3x=-21\ \ |:(-3)\\\\x=7[/tex]
[tex]5b)\\\\ 4( \frac{1}{2} x-2)^2<(1-x)^2 + 7\\\\4(\frac{1}{4}x^2-2x+4)<1-2x+x^2+7\\\\x^2-8x+16<-2x+x^2+8\\\\x^2-8x+2x-x^2 <8-16\\\\ -6x<-8\ \ |:(-6)\\\\x>\frac{8}{6}\\\\x>\frac{4}{3}[/tex]
