Question

Resolva as equações do 2°grau a seguir:

a)x²-2x-15=0
b)x²-3x-40=0
c)x²-4x-12=0
d)x²-5x-50=0​

Answer 1

Resposta:

$a) \: \: \: \: \: {x}^{1} = 5 \\ {x}^{2} = - 3 \\ \\ b) \: \: \: \: \: {x}^{1} = 8\\ {x}^{2} = - 5\\ \\ c) \: \: \: \: \: {x}^{1} = 6\\ {x}^{2} = - 2\\ \\ d) \: \: \: \: \: {x}^{1} = 20 \\ {x}^{2} =5$

Explicação passo-a-passo:

$a) \: \: \: \: \: {x}^{2} - 2x - 15 = 0 \\ \\ a = 1 \\ b = - 2 \\ c = - 15 \\ \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x = \frac{ - ( - 2) + - \sqrt{ { ( - 2)}^{2} - 4 \times 1 \times ( - 15)} }{2 \times 1} \\ x = \frac{2 + - \sqrt{4 + 60} }{2} \\ x = \frac{2 + - \sqrt{64} }{2} \\ x = \frac{2 + - 8}{2} \\ \\ {x}^{1} = \frac{2 + 8}{2} = {x}^{1} = \frac{10}{2} = {x}^{1} = 5 \\ {x}^{2} = \frac{2 - 8}{2} = {x}^{2} = - \frac{6}{2} = {x}^{2} = - 3$

$b) \: \: \: \: \: {x}^{2} - 3x - 40 \\ \\ a = 1 \\ b = - 3 \\ c = - 40 \\ \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x = \frac{ - ( - 3) + - \sqrt{ {( - 3)}^{2} - 4 \times 1 \times ( - 40)} }{2 \times 1} \\ x = \frac{3 + - \sqrt{9 + 160} }{2} \\ x = \frac{3 + - \sqrt{169} }{2 } \\ x = \frac{3 + - 13}{2} \\ \\ {x}^{1} = \frac{3 + 13}{2} = {x}^{1} = \frac{16}{2} = {x}^{1} = 8 \\ {x }^{2} = \frac{3 - 13}{2} = {x}^{2} = - \frac{10}{2} = {x}^{2} = - 5$

$c) \: \: \: \: \: {x}^{2} - 4x - 12 = 0 \\ \\ a = 1 \\ b = - 4 \\ c = - 12 \\ \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x = \frac{ - ( - 4) + - \sqrt{ {( - 4)}^{2} - 4 \times 1 \times ( - 12)} }{2 \times 1} \\ x = \frac{4 + - \sqrt{16 + 48} }{2 } \\ x = \frac{4 + - \sqrt{64} }{2} \\ x = \frac{4 + - 8}{2} \\ \\ {x}^{1} = \frac{4 + 8}{2} = {x}^{1} = \frac{12}{2} = {x}^{1} = 6 \\ {x}^{2} = \frac{4 - 8}{2} = {x}^{2} = - \frac{4}{2} = {x}^{2} = - 2$

$d) \: \: \: \: \: {x}^{2} - 5x - 50 = 0 \\ \\ a = 1 \\ b = - 5 \\ c = - 50 \\ \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x = \frac{ - ( - 5) + - \sqrt{ {( - 5)}^{2} - 4 \times 1 \times ( - 50) } }{2 \times 1} \\ x = \frac{25 + - \sqrt{25 + 200} }{2} \\ x = \frac{25 + - \sqrt{225} }{2} \\ x = \frac{25 + - 15}{2} \\ \\ {x}^{1} = \frac{25 + 15}{2} = {x}^{1} = \frac{40}{2} = {x}^{1} = 20 \\ {x}^{2} = \frac{25 - 15}{2} = {x}^{2} = \frac{10}{2} = {x}^{2} = 5$