Question

Me ajudem pfv preciso urgente​

Answer 1

$1. \: a) {x}^{2} - 5x + 6 = 0 \\ a = 1 \\ b = - 5 \\ c = 6 \\ \\ x = \frac{ - ( - 5) + - \sqrt{ {( - 5)}^{2} - 4 \times 1 \times 6 } }{2 \times 1} \\ x = \frac{5 + - \sqrt{25 - 24} }{2} \\ x = \frac{5 + - \sqrt{1} }{2} \\ {x}^{1} = \frac{5 + 1}{2} = \frac{6}{2} = \frac{6 \div 2}{2 \div 2} = \frac{3}{1} = 3 \\ {x}^{2} = \frac{5 - 1}{2} = \frac{4}{2} = \frac{4 \div 2}{ 2\div 2} = \frac{2}{1} = 2$

$b) {4x}^{2} - 8x + 4 \\ a = 4 \\ b = - 8 \\ c = 4 \\ x = \frac{ - ( - 8) + - \sqrt{ {( - 8)}^{2} - 4 \times 4 \times 4 } }{2 \times 4} \\ x = \frac{8 + - \sqrt{64 - 64} }{8} \\ x = \frac{8 + - \sqrt{0} }{8} \\ x = \frac{8}{8} \\ x = \frac{8 \div 8}{8 \div 8} \\ x = \frac{1}{1} \\ x = 1$

$c) {x}^{2} + 2x - 8 = 0 \\ a = 1 \\ b = 2 \\ c = - 8 \\ x = \frac{ -2 + - \sqrt{ {2}^{2} - 4 \times 1 \times ( - 8) } }{2 \times 1} \\ x = \frac{ - 2 + - \sqrt{4 + 32} }{2} \\ x = \frac{ - 2 + - \sqrt{36} }{2} \\ {x}^{1} = \frac{ - 2 + 6}{2} = \frac{4}{2} = 2 \\ {x}^{2} = \frac{ - 2 - 6}{2} = - \frac{8}{2} = - 4$

$d) {x}^{2} - 5x + 8 = 0 \\ a = 1 \\ b = - 5 \\ c = 8 \\ x = \frac{ - ( - 5) + - \sqrt{ {( - 5)}^{2} - 4 \times 1 \times 8 } }{2 \times 1} \\ x = \frac{5 + - \sqrt{25 - 32} }{2} \\ x = \frac{5 + - \sqrt{ - 7} }{2}$