Question

Determine o conjunto solução das equações utilizando a formula de Bhaskara
a) x²-7x+6=0
b) 9x²-6+1=0
c) 6x²+x-1=0
d) 5x²-3x-2=0
e) x²+3x-4=0

Answer 1

Formula de Bhaskara:

$x = \frac{-b \pm \sqrt{\Delta}}{2*a} \\ \\ \Delta = b^2 - 4 * a * c$

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a) $x^{2} -7x + 6 = 0 \\ \\ a = 1,\, b = -7,\, c = 6 \\ \\ \Delta = (-7)^2-4*1*6 \\ \Delta = 49 - 24 \\ \Delta = 25 \\ \\ x = \frac{-(-7)\pm\sqrt{25}}{2*1} \\ \\ x= \frac{7\pm5}{2} \\ \\ x' = \frac{7+5}{2} = \frac{12}{2} = 6 \\ \\ x'' = \frac{7-5}{2} = \frac{2}{2} = 1$

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b) $9x^2 -6x + 1 = 0 \\ \\ a=9,\,b=-6,\,c=1 \\ \\ \Delta = (-6)^2-4*9*1 \\ \Delta=36-36 \\ \Delta = 0 \\ \\ x= \frac{-(-6)\pm\sqrt{0}}{2*9} \\ \\ x=\frac{6\pm0}{18} \\ \\ x'=x''= \frac{6}{18} = \frac{1}{3}$

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c) $6x^2 +x-1=0 \\ \\ a=6,\,b=1,\,c=-1 \\ \\ \Delta=1^2-4*6*(-1) \\ \Delta = 1+24 \\ \Delta =25 \\ \\ x=\frac{-1\pm\sqrt{25}}{2*6} \\ \\ x= \frac{-1\pm5}{12} \\ \\ x'= \frac{-1+5}{12} = \frac{4}{12} = \frac{1}{3} \\ \\ x''= \frac{-1-5}{12} = \frac{-6}{12} = - \frac{1}{2} = -0,5$

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d) $5x^2-3x-2 = 0 \\ \\ a=5,\,b=-3,\,c=-2 \\ \\ \Delta=(-3)^2-4*5*(-2) \\ \Delta=9+40 \\ \Delta=49 \\ \\ x= \frac{-(-3)\pm\sqrt{49}}{2*5} \\ \\ x= \frac{3\pm7}{10} \\ \\ x'= \frac{3+7}{10} = \frac{10}{10} = 1 \\ \\ x''= \frac{3-7}{10} = \frac{-4}{10} = -0,4$

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e) $x^{2} +3x-4=0 \\ \\ a=1,\,b=3,\,c=-4 \\ \\ \Delta=3^2-4*1*(-4) \\ \Delta=9+16 \\ \Delta=25 \\ \\ x= \frac{-3\pm\sqrt{25}}{2*1} \\ \\ x= \frac{-3\pm5}{2} \\ \\ x'= \frac{-3+5}{2} = \frac{2}{2} =1 \\ \\ x''= \frac{-3-5}{2} = \frac{-8}{2} =-4$