Question

1. Utilizando o processo algébrico de
Bhaskara, determine as raízes das equações
do 2 grau no conjunto dos números reais:
a) x2 + 4x - 5 = 0
b) 2x? - 9x + 4 = 0
c) x + 8x + 16 = 0​

Answer 1

Resposta:

Bhaskara

Explicação passo-a-passo:

a)

$x^{2} +4x-5=0\\a=1 \\b= 4\\c= -5\\\\\frac{-4 \frac{+}{}\sqrt{4^{2}-4.(-5).1 } }{2.1} \\ \frac{-4 \frac{+}{}\sqrt{16 +20 } }{2} \\ \\\frac{-4 \frac{+}{}\sqrt{36 } }{2}\\\\ \frac{-4 \frac{+}{} 6}{2} \\\\ x_{1}= \frac{-4 + 6}{2} = \frac{2}{2} = 1\\\\ x_{2}= \frac{-4 - 6}{2} = \frac{-10}{2} = -5\\\\\\x_{1}= 1 \\\\x_{2}=-5$

b)

$2x^{2} -9x+4=0\\a=2 \\b= -9\\c= 4\\\\\frac{-(-9) \frac{+}{}\sqrt{(-9)^{2}-4.4.2 } }{2.2} \\ \\ \frac{9 \frac{+}{}\sqrt{81-32 } }{4} \\ \\\frac{9 \frac{+}{}\sqrt{ 49} }{4}\\\\ \frac{9 \frac{+}{} 7}{4} \\\\ x_{1}= \frac{9 + 7}{4} = \frac{16}{4} = 4\\\\ x_{2}= \frac{9 - 7}{4} = \frac{2}{4} = \frac{1}{2}\\\\\\x_{1}= 4 \\\\x_{2}=\frac{1}{2}$

c)

$x^{2} +8x+16=0\\a=1 \\b= 8\\c= 16\\\\\frac{-8 \frac{+}{}\sqrt{8^{2}-4.16.1 } }{2.1} \\ \\ \frac{-8 \frac{+}{}\sqrt{64-64 } }{2} \\ \\\frac{-8 \frac{+}{}\sqrt{ 0} }{2}\\\\ \frac{-8 \frac{+}{} 0}{2} \\\\ x_{1}= \frac{-8+0}{2} = - 4\\\\ x_{2}= \frac{-8-0}{2} =-4 \\\\x_{1}= -4 \\\\x_{2}=-4$