Question

Ajudem ai 15 pontos​

Answer 1

Explicação passo-a-passo:

Seja a equação de bhaskara

$x=\frac{-b+-\sqrt{b^2-4.a.c} }{2.a}$

$a)x^2-3x+2\\a=1\\b=-3\\c=2\\\\x=\frac{3+-\sqrt{9-8} }{2}\\x=\frac{3+-\sqrt{1} }{2}\\\\x_1=\frac{3+1}{2}=\frac{4}{2} \\x_1=2\\x_2=\frac{3-1}{2}=\frac{2}{2} \\x_2=1$

a) - IV)

$b)y^2-7y+12=0$

$a=1\\b=-7\\c=12$

$x=\frac{7+-\sqrt{49-48} }{2}\\x_1=\frac{7+1}{2}=\frac{8}{2} \\x_1=4\\x_2=\frac{7-1}{2}=\frac{6}{2} \\x_2=3$

b) - I)

$c)x^2-5x-6=0\\a=1\\b=-5\\c=-6$

$x=\frac{5+-\sqrt{25+24} }{2}\\x=\frac{5+-\sqrt{49} }{2}\\x_1=\frac{5+7}{2}=\frac{12}{2} \\x_1=6\\x_2=\frac{5-7}{2}=\frac{-2}{2} \\x_2=-1$

c) - II)

$d)t^2+6t+8=0\\a=1\\b=6\\c=8$

$x=\frac{-6+-\sqrt{36-32} }{2}\\x=\frac{-6+-\sqrt{4} }{2}\\\\x_1=\frac{-6+2}{2}=\frac{-4}{2} \\x_1=-2\\x_2=\frac{-6-2}{2}=\frac{-8}{2} \\x_2=-4$

d) - III)