Question

calcule as equações abaixo.
Com a fórmula de Bhaskara.
a) x² + 4x - 5 = 0
b) x² + 8x + 16 = 0
c) x² - 3x - 28 = 0
d) x² -x-1=0
e) x² - 20x + 51 = 0​

Answer 1

Resposta:

a) $x_{1}$ = 1          $x_{2}$ = -5

b) $x$ = -4

c) $x_{1}$ = 7          

d)

e)

Explicação passo-a-passo:

a) $x^{2} +4x-5=0\\\\a=1\\b=4\\c=-5\\\\\frac{-b+-\sqrt{b^{2}-4ac } }{2a} =\\\frac{-4+-\sqrt{4^{2}-4.1.(-5) } }{2.1}= \\\frac{-4+-\sqrt{16-4.(-5)} }{2}=\\\frac{-4+-\sqrt{16+20} }{2}=\\ \frac{-4+-\sqrt{36} }{2}=\\ \frac{-4+-6}{2} =\\-2+-3=\\\\x_{1}=-2+3 >x_{1}=1\\x_{2}=-2-3>x_{2}=-5$

b) $x^{2} +8x+16=0\\\\a=1\\b=8\\c=16\\\\\frac{-b+-\sqrt{b^{2}-4ac } }{2a}=\\ \frac{-8+-\sqrt{8^{2}-4.1.16 } }{2.1}=\\ \frac{-8+-\sqrt{64-4.16} }{2}=\\ \frac{-8+-\sqrt{64-64} }{2}=\\ \frac{-8+-\sqrt{0} }{2}=\\ \frac{-8+-0}{2}=\\ \frac{-8}{2}=\\ \\x=-4$

c) $x^{2} -3x-28=0\\\\a=1\\b=-3\\c=-28\\\\\frac{-b+-\sqrt{b^{2}-4ac } }{2a}=\\ \frac{-(-3)+-\sqrt{(-3)^{2}-4.1.(-28) } }{2}=\\ \frac{3+-\sqrt{9-4.(-28)} }{2}=\\\frac{3+-\sqrt{9+112} }{2}=\\ \frac{3+-\sqrt{121} }{2}=\\ \frac{3+-11}{2}= \\\\x_{1}=\frac{3+11}{2}>x_{1}=\frac{14}{2}>x_{1}=7\\x_{2}=\frac{3-11}{2}>x_{2}=\frac{-8}{2}>x_{2}=-4$

d)$x^{2} -x-1=0\\\\a=1\\b=-1\\c=-1\\\\\frac{-b+-\sqrt{b^{2}-4ac } }{2a}=\\ \frac{-(-1)+-\sqrt{(-1)^{2}-4.1.(-1)} }{2.1}=\\\frac{1+-\sqrt{1-4.(-1)} }{2}=\\ \frac{1+-\sqrt{1+5} }{2}=\\ \frac{1+-\sqrt{6} }{2}=$

vamos fatorar o 6 e continuar a equação

$6|2\\3|3\\\\2\sqrt{3}$

$\frac{1+-\sqrt{6} }{2}=\\ \frac{1+-2\sqrt{3} }{2}=\\\\x_{1}=\frac{1+2\sqrt{3} }{2}>x_{1}=\frac{3\sqrt{3} }{2}\\ x_{2}=\frac{1-2\sqrt{3} }{2}>x_{2}=\frac{-\sqrt{3} }{2}$

e) $x^{2} -20x+51=0\\\\a=1\\b=-20\\c=51\\\\\frac{-b+-\sqrt{b^{2}-4ac } }{2a}=\\\frac{-(-20)+-\sqrt{(-20)^{2}-4.1.51 } }{2}=\\ \frac{20+-\sqrt{400-4.51}}{2}=\\ \frac{20+-\sqrt{400-204} }{2}=\\ \frac{20+-\sqrt{196} }{2}=\\\frac{20+-14}{2}=\\ 10+-7=\\\\x_{1}=10+7>x_{1}=17\\x_{2}=10-7>x_{2}=3$