Question

Determine o valor de x no triângulo retângulo abaixo usando o Teorema de Pitágoras

Answer 1

Resposta:

a)

${x}^{2} - 5x - 4 = 0$

$∆ = { - 5}^{2} - 4 \times 1 \times 6 \\ ∆ = 25 - 4 \times 6 \\ ∆ = 25 - 24 = 1$

$x = \frac{ - ( - 5) + \sqrt{1} }{2 \times 1} \\ x = \frac{5 + 1}{2} = \frac{6}{2} = 3$

$x" = \frac{ - ( - 5) - \sqrt{1} }{2 \times 1} \\ x" = \frac{5 - 1}{2} = \frac{4}{2} = 2$

b)

${x}^{2} - 8x + 12 = 0$

$∆ = { - 8}^{2} - 4 \times 1 \times 12 \\ ∆ = 64 - 4 \times 12 \\ ∆ = 64 - 48 \\∆ =16$

$x = \frac{ - ( - 8) + \sqrt{16} }{2 \times 1} \\ x = \frac{8 + 4}{2} = \frac{12}{2} = 6$

$x" = \frac{ - ( - 8) - \sqrt{16} }{2 \times 1} \\ x" = \frac{8 - 4}{2} = \frac{4}{2} = 2$

c)

${x}^{2} - 3x - 4 = 0$

$∆ = { - 3}^{2} - 4 \times 1 \times - 4 \\ ∆ = 9 - 4 \times - 4 \\∆ = 9 + 16 \\ ∆ = 25$

$x = \frac{ - ( - 3) + \sqrt{25} }{2 \times 1} \\ x = \frac{3 + 5}{2} = \frac{8}{2} = 4$

$x" = \frac{ - ( - 3) - \sqrt{25} }{2 \times 1} \\x" = \frac{3 - 5}{2} = \frac{ - 2}{2} = - 1$

d)

${x}^{2} - 2x - 8 = 0$

$∆ = { - 2}^{2} - 4 \times 1 \times - 8 \\ ∆ = 4 - 4 \times - 8 \\ ∆ = 4 + 32 \\ ∆ = 36$

$x = \frac{ - ( - 2) + \sqrt{36} }{2 \times 1} \\ x = \frac{2 + 6}{2} = \frac{8}{2} = 4$

$x" = \frac{ - ( - 2) - \sqrt{36} }{2 \times 1} \\ x" = \frac{2 - 6}{2} = \frac{ - 4}{2} = - 2$

espero ter ajudado!