I)x-6=-12x²j)x(4x+20) +25=0.... Pergunta de ideia deStefhany39

May 2019 2 94 Report

I)x-6=-12x²
j)x(4x+20) +25=0

mithie7552

Resposta:

$i)\\ x-6=-12x^2\\ 12x^2+x-6=0\\ \\ a=12\\ b=1\\ c=-6\\ \\ \Delta=b^2-4ac\\ \Delta=1^2-4(12)(-6)\\ \Delta=1+288\\ \Delta=289\\ \\ x={-1\pm\sqrt{289} \over2(12)}={-1\pm17\over24}\\ \\ x'={-1+17\over24}={16\over24}={2\over3}\\ \\ x"={-1-17\over24}=-{18\over24}=-{3\over4}\\ \\ S=\{-{3\over4},{2\over3}\}$

$j)\\ x(4x+20)+25=0\\ 4x^2+20x+25=0\\ \\ a=4\\ b=20\\ c=25\\ \\ \Delta=b^2-4ac\\ \Delta=400-400\\ \Delta=0\\ \\ x'=x"={-20\pm\sqrt{0} \over2.(4)}=-{20\over8}=-{5\over2}\\ \\ S=\{-{5\over2}\}$