Question

usandoa formula de baskara resolva ) x2 — 10x + 24 = 0; c) x2— 4x+12 = 0 ; d) x2_8x+8=0; e) x2-6x+9=0;
ajudaaaa pfvr

Answer 1

Explicação passo-a-passo:

$b) \: \: \: {x}^{2} - 10x + 24 = 0\\ \\ a = 1 \\ b = - 10 \\ c = 24 \\ \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} \\ x = \frac{ - ( - 10) + - \sqrt{ {( - 10)}^{2} - 4 \times 1 \times 24} }{2 \times 1} \\ x = \frac{10 + - \sqrt{100 - 96} }{2} \\ x = \frac{10 + - \sqrt{4} }{2} \\ x = \frac{10 + - 2}{2} \\ \\ {x}^{1} = \frac{10 + 2}{2} = {x}^{1} = \frac{12}{2} = {x}^{1} = 6 \\ {x}^{2} = \frac{10 - 2}{2} = {x}^{2} = \frac{8}{2} = {x}^{2} = 4$

$d) \: \: \: {x}^{2} - 8x + 8 = 0 \\ \\ a = 1 \\ b = - 8 \\ c = 8 \\ \\ x = \frac{ - ( - 8) + - \sqrt{ {( - 8)}^{2} - 4 \times 1 \times 8} }{2 \times 1} \\ x = \frac{8 + - \sqrt{64 - 32} }{2} \\ x = \frac{8 + - \sqrt{32} }{2} \\ x = \frac{8 + - 4 \sqrt{2} }{2}$

$e) \: \: \: {x}^{2} - 6x + 9 \\ \\ a = 1 \\ b = - 6 \\ c = 9 \\ \\ x = \frac{ - ( - 6) + - \sqrt{ {( - 6)}^{2} - 4 \times 1 \times 9 } }{2 \times 1} \\ x = \frac{6 + - \sqrt{36 - 36} }{2} \\ x = \frac{6 + - 0}{2} = x = \frac{6}{2} = x = 3$

$c) \: \: \: {x}^{2} - 4x + 12 = 0\\ \\ a = 1 \\ b = - 4 \\ c = 12 \\ \\ x = \frac{ - ( - 4) + - \sqrt{ {( - 4)}^{2} - 4 \times 1 \times 12 } }{2 \times 1} \\ x = \frac{4 + - \sqrt{16 - 48} }{2} \\ x = \frac{4 + - \sqrt{ - 32} }{2}$