Question

me ajudem pfvr alguém?????

Answer 1

Resposta:

Explicação passo a passo:

$a) x^{2}-5x+6=0\\ a=1\\b=-5\\c=6\\x=\frac{-(-5)+-\sqrt{(-5)^{2} -4.1.6} }{2.1} \\\\x=\frac{5+-\sqrt{25-24} }{2} \\x=\frac{5+-1}{2} \\x1=\frac{5+1}{2}=3 \\x2=\frac{5-1}{2} =2\\\\x1=3, x2=2$        $b) x^{2}-8x+12=0\\ a=1\\b=-8\\c=12\\x=\frac{-(-8)+-\sqrt{(-8)^{2} -4.1.12} }{2.1} \\\\x=\frac{8+-\sqrt{64-48} }{2} \\x=\frac{8+-4}{2} \\x1=\frac{8+4}{2}=6 \\x2=\frac{8-4}{2} =2\\\\x1=6, x2=2$  $c) x^{2}+2x-8=0\\ a=1\\b=2\\c=-8\\x=\frac{-2+-\sqrt{2^{2} -4.1.(-8)} }{2.1} \\\\x=\frac{-2+-\sqrt{4+32} }{2} \\x=\frac{-2+-6}{2} \\x1=\frac{-2+6}{2}=2 \\x2=\frac{-2-6}{2} =-4\\\\x1=2, x2=-4$      

$d)6 x^{2}+x-1=0\\ a=6\\b=1\\c=-1\\x=\frac{-1+-\sqrt{1^{2} -4.6.(-1)} }{2.6} \\\\x=\frac{-1+-\sqrt{1+24} }{12} \\x=\frac{-1+-5}{12} \\x1=\frac{-1+5}{12}=\frac{1}{3} \\x2=\frac{-1-5}{12} =-\frac{1}{2} \\\\x1=\frac{1}{3} , x2=-\frac{1}{2}$    $e) 3x^{2}-7x+2=0\\ a=3\\b=-7\\c=2\\x=\frac{-(-7)+-\sqrt{(-7)^{2} -4.3.2} }{2.3} \\\\x=\frac{7+-\sqrt{49-24} }{6} \\x=\frac{7+-5}{6} \\x1=\frac{7+5}{6}=2 \\x2=\frac{7-5}{6} =\frac{1}{3} \\\\x1=2, x2=\frac{1}{3}$     $f) 2x^{2}-7x=15\\\\ 2x^{2}-7x-15=0\\a=2\\b=-7\\c=-15\\x=\frac{-(-7)+-\sqrt{(-7)^{2} -4.2.(-15)} }{2.2} \\\\x=\frac{7+-\sqrt{49+120} }{4} \\x=\frac{7+-13}{4} \\x1=\frac{7+13}{4}=5 \\x2=\frac{7-5}{4} =-\frac{3}{2} \\\\x1=5, x2=-\frac{3}{2}$