Matemática, perguntado por Rafael1203i, 10 meses atrás

Resolva as equações dos 2º grau em R, por favor

a)x²-5x+5=0

b)x²-8x+12=0

c)x²+2x-8=0

d)x²-5x+8=0

e)2x²-8x+8=0

f)x²-4x-5=0


Soluções para a tarefa

Respondido por crmacena
2

Resposta:

a)x²-5x+5=0

a = 1

b = -5

c = 5

x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-5) +- \sqrt{(-5)^{2} - 4 * 1 * 5} }{2*1}\\\\x = \frac{5+-\sqrt{25-20} }{2}\\\\x = \frac{5+-\sqrt{5} }{2}\\\\x' = \frac{5+\sqrt{5} }{2}\\\\x'' = \frac{5-\sqrt{5} }{2}

b)x²-8x+12=0

a = 1

b = -8

c = 12

x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-8) +- \sqrt{(-8)^{2} - 4 * 1 * 12} }{2*1}\\\\x = \frac{8+-\sqrt{64-48} }{2}\\\\x = \frac{8+-\sqrt{16} }{2}\\\\x = \frac{8+-4}{2}\\\\x' = \frac{8+4}{2}=\frac{12}{2}=6 \\\\x'' = \frac{8-4}{2}=\frac{4}{2} =2

c)x²+2x-8=0

a = 1

b = 2

c = -8

x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(2) +- \sqrt{(2)^{2} - 4 * 1 * -8} }{2*1}\\\\x = \frac{-2+-\sqrt{4+32} }{2}\\\\x = \frac{-2+-\sqrt{+36} }{2}\\\\x = \frac{-2+-6}{2}\\\\x' = \frac{-2+6}{2}=\frac{4}{2}=2 \\\\x'' = \frac{-2-6}{2}=\frac{-8}{2} =-4

d)x²-5x+8=0

a = 1

b = -5

c = 8

x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-5) +- \sqrt{(-5)^{2} - 4 * 1 * 8} }{2*1}\\\\x = \frac{5+-\sqrt{25-32} }{2}\\\\x = \frac{5+-\sqrt{-7} }{2}\\\\x' = \frac{5+\sqrt{-7} }{2}\\\\x'' = \frac{5-\sqrt{-7} }{2}\\\\

e)2x²-8x+8=0

a = 2

b = -8

c = 8

x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-8) +- \sqrt{(-8)^{2} - 4 * 2 * 8} }{2*2}\\\\x = \frac{8+-\sqrt{64-64} }{4}\\\\x = \frac{8+-\sqrt{0} }{4}\\\\x = \frac{8+-0}{4}\\\\x' = \frac{8+0}{4}=\frac{8}{4}= 2\\\\x'' = \frac{8-0}{4}=\frac{8}{4}=2

f)x²-4x-5=0

a = 1

b = -4

c = -5

x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-4) +- \sqrt{(-4)^{2} - 4 * 1 * -5} }{2*1}\\\\x = \frac{4+-\sqrt{16+20} }{2}\\\\x = \frac{4+-\sqrt{36} }{2}\\\\x = \frac{4+-6}{2}\\\\x' = \frac{4+6}{2}=\frac{10}{2} =5\\\\x'' = \frac{4-6}{2}=\frac{-2}{2} =-1

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