Question

alguém poderia me ajudar por favor? preciso pra hoje​

Answer 1

Resposta:

Explicação passo-a-passo:

I.  x² - 5x - 6 = 0

$x^{2} -5x-6=0\\\\delta=b^{2} -4ac\\delta=(-5)^{2} -4.1.(-6)\\delta=25+24\\delta=49\\\\x'=\frac{-b+\sqrt{delta} }{2a} \\\\x'=\frac{5+7}{2} \\\\x'=\frac{12}{2} \\\\x'=6\\\\\\x"=\frac{-b-\sqrt{delta} }{2a} \\\\x"=\frac{5-7}{2} \\\\x"=\frac{-2}{2} \\\\x"=-1$

S = { -1, 6 }

II. x² - 8x + 16 = 0

$x^{2} -8x+16=0\\\\delta=b^{2} -4ac\\delta=(-8)^{2} -4.1.16\\delta=64-64\\delta=0\\\\x'=\frac{-b+\sqrt{delta} }{2a} \\\\x'=\frac{8+0}{2} \\\\x'=\frac{8}{2} \\\\x'=4\\\\\\x"=\frac{-b-\sqrt{delta} }{2a} \\\\x"=\frac{8-0}{2} \\\\x"=\frac{8}{2} \\\\x"=4$

S = { 4 }

III. x² - 7x + 10 = 0

$x^{2}-7x+10=0 \\\\delta=b^{2} -4ac\\delta=(-7)^{2} -4.1.10\\delta=49-40\\delta=9\\\\x'=\frac{-b+\sqrt{delta} }{2a} \\\\x'=\frac{7+3}{2} \\\\x'=\frac{10}{2} \\\\x'=5\\\\\\x"=\frac{-b-\sqrt{delta} }{2a} \\\\x"=\frac{7-3}{2} \\\\x"=\frac{4}{2} \\\\x"=2$

S = { 2, 5 }

IV. x² - x - 2 = 0

$x^{2} -x-2=0\\\\delta=b^{2} -4ac\\delta=(-1)^{2} -4.1.(-2)\\delta=1+8\\delta=9\\\\x'=\frac{-b+\sqrt{delta} }{2a} \\\\x'=\frac{1+3}{2} \\\\x'=\frac{4}{2} \\\\x'=2\\\\\\x"=\frac{-b-\sqrt{delta} }{2a} \\\\x"=\frac{1-3}{2} \\\\x"=\frac{-2}{2} \\\\x"=-1$

S = { -1, 2 }

V. 2x² + 7x + 3 = 0

$2x^{2}+7x+3=0 \\\\delta=b^{2} -4ac\\delta=7^{2} -4.2.3\\delta=49-24\\delta=25\\\\x'=\frac{-b+\sqrt{delta} }{2a} \\\\x'=\frac{-7+5}{4} \\\\x'=\frac{-2}{4} \\\\x'=-\frac{1}{2} \\\\\\x"=\frac{-b-\sqrt{delta} }{2a} \\\\x"=\frac{-7-5}{4} \\\\x"=\frac{-12}{4} \\\\x"=-3$

S = { -3, -1/2 }

VI. x² - 12x + 35 = 0

$x^{2} -12x+35=0\\\\delta=b^{2} -4ac\\delta=(-12)x^{2} -4.1.(35)\\delta=144-140\\delta=4\\\\x'=\frac{-b+\sqrt{delta} }{2a} \\\\x'=\frac{12+2}{2} \\\\x'=\frac{14}{2} \\\\x'=7\\\\\\x"=\frac{-b-\sqrt{delta} }{2a} \\\\x"=\frac{12-2}{2} \\\\x"=\frac{10}{2} \\\\x"=5$

S = { 5, 7 }

Espero ter ajudado!

Desculpe qualquer erro.