Question

por favor ajuda neste problema

Answer 1

3)

a)

${x}^{2} = 16.12 \\ {x}^{2} = 192 \\ x = \sqrt{192} \\ x = 8 \sqrt{3}$

b)

${12}^{2} = 24.(24 - x) \\ 24 - x = \frac{144}{24} \\ 24 - x = 6 \\ x = 24 - 6 \\ x = 18$

c)

${x}^{2} = 9.5 \\ x = \sqrt{9.5} = 3 \sqrt{5}$

d)

${4}^{2} = x(x + 6) \\{x}^{2} + 6x = 16 \\ {x}^{2} + 6x - 16 = 0$

$Δ = {b}^{2} - 4ac \\ Δ = {6}^{2} - 4.1.( - 16) \\ Δ = 36 + 64 \\ Δ = 100$

$x = \frac{ - b ± \sqrt{Δ} }{2a} \\ x = \frac{ - (6) ± \sqrt{100} }{2.1} \\ x = \frac{ - 6 ±10 }{2}$

$x' = \frac{ - 6 + 10}{2} = \frac{4}{2} = 2$

e)

${10}^{2} =( 2x + 5).2x \\ x(2x + 5) = \frac{100}{2} \\ 2 {x}^{2} + 10x = 50 \div (2) \\ {x}^{2} + 5x = 25$

${x}^{2} + 5x - 25 = 0</p><p> \\ Δ = 25 + 100 = 125 \\ x = \frac{ - 5 + \sqrt{125} }{2} = \frac{ - 5 + 5 \sqrt{5} }{2}$