Question

resolva as equações
b) x2-5x + 8 = 0
c) x² - 4x +9=0​

Answer 1

Resposta:

$x1 = ( - b + \sqrt{b {}^{2} - 4ac } ) \div 2a \\ x1 = ( - ( - 5) + \sqrt{25 - 4 \times 1 \times 8} ) \div 2 \\ x1 = (5 + \sqrt{25 - 32} ) \div 2 \\ x1 =( 5 + \sqrt{ - 7} ) \div 2 \\ x1 = \frac{5 + \sqrt{ - 1 \times 7} }{2} \\ x1 = \frac{5 + \sqrt{i {}^{2} \times 7 } }{2} \\ x1 = \frac{5 + i \sqrt{7} }{2} \\ x2 =( 5 - i \sqrt{7} ) \div 2$

$x {}^{2} - 4x + 9 = 0 \\ x1 = \frac{ - b + \sqrt{b {}^{2} - 4ac } }{2a} \\ x1 = (4 + \sqrt{16 - 4 \times 1 \times 9} ) \div 2 \times 1 \\ x1 = \frac{4 + \sqrt{16 - 36} }{2} \\ x1 = \frac{4 + \sqrt{ - 20} }{2} \\ x1 = \frac{4 + \sqrt{20i {}^{2} } }{2} \\ x1 = \frac{4 + i \sqrt{20} }{2} \\ x1 = \frac{4 + i \sqrt{2 {}^{2} \times 5 } }{2} \\ x1 = \frac{4 + 2i \sqrt{5} }{2} \\ x1 = \frac{2(2 + i \sqrt{5}) }{2} \\ x1 = 2 + i \sqrt{5} \\ x2 = 2 - i \sqrt{5}$