Question

sistema equação de 2 grau a) y=8-x x²+y²=40 b) x/y=1/4 x²+y²=153

Answer 1

Resposta:

a) {6; 2) ou {2; 6)

b) (3; 12) ou (-3; -12)

Explicação passo-a-passo:

a)

y = 8 - x

x² + y² = 40

x² + (8 - x)² = 40

x² + 64 - 16x + x² = 40

2x² - 16x + 24 = 0

x² - 8x + 12 = 0

Δ = 64 - 48 = 16

x = (8 ± 4) / 2

x = 6                                    x = 2

y = 8 - 6 = 2                         y = 8 - 2 = 6

b)

b) x / y = 1/4

y = 4x

x² + y² = 153

x² + 16x² = 153

17x² = 153

x² = 9

x = 3          y = 12

ou

x = -3        y = -12

Answer 2

Oie, Td Bom?!

a)

$\left\{\begin{matrix}</p><p>y= 8- x& & \\ </p><p> x {}^{2} + y {}^{2} = 40 & & </p><p>\end{matrix}\right.$

$x {}^{2} + (8 - x) {}^{2} = 40$

$x {}^{2} + 64 - 16x + x {}^{2} = 40$

$2x {}^{2} + 64 - 16x = 40$

$2x {}^{2} + 64 - 16x - 40 = 0$

$2x {}^{2} + 24 - 16x = 0$

$2x {}^{2} - 16x + 24 = 0$

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

$x {}^{2} - 2x - 6x + 12 = 0$

$x \: . \: (x - 2) - 6(x - 2) = 0$

$(x - 2) \: . \: (x - 6) = 0$

$x - 2 = 0 \\ x - 6 = 0$

$x = 2 \\ x = 6$

$y = 8 - 2 \\ y = 8 - 6$

$y = 6 \\ y = 2$

$(x _{1} \: , \: y_{1}) = (2 \: , \: 6) \\ (x _{2} \: , \: y _{2}) = (6 \: , \: 2)$

b)

$\left\{\begin{matrix}\frac{x}{y} = \frac{1}{4} & & \\ </p><p>x {}^{2} + y {}^{2} = 153 & & </p><p>\end{matrix}\right.$

$\left\{\begin{matrix}</p><p> 4x = y& & \\ </p><p> x {}^{2} + y {}^{2} = 153 & & </p><p>\end{matrix}\right.$

$x {}^{2} + (4x) {}^{2} = 153$

$x {}^{2} + 16x {}^{2} = 153$

$17x {}^{2} = 153$

$x {}^{2} = 9$

$x = ±3$

$x = - 3 \\ x = 3$

$4 \: . \: ( - 3) = y \\ 4 \: . \: 3 = y$

$y = - 12 \\ y = 12$

$(x _{1} \: , \: y _{1} ) = ( - 3 \: , \: - 12) \\ (x _{2} \: , \: y _{2}) = (3 \: , \: 12)$

Att. Makaveli1996