Question

me ajuda por favor ;-; ​

Answer 1

a)

$2 {(x}^{2} - 1) = {x}^{2} + 7 \\ 2 {x}^{2} - 2 = {x}^{2} + 7 \\ 2 {x}^{2} - {x}^{2} = 2 + 7 \\ {x}^{2} = 9$

$x = ± \sqrt{9} = ±3$

s={-3,3}

b)

$6.( {x}^{2} - 1) = 5 {x}^{2} + 5 \\ 6 {x}^{2} - 6 = 5 {x}^{2} + 5 \\ 6 {x}^{2} - 5 {x}^{2} = 5 + 6 \\ {x }^{2} = 11$

$x = ± \sqrt{11}$

s={-√11,√11}

b)

$(x - 4)(x +3) = - 12 \\ {x}^{2} + 3x - 4x - 12 = - 12 \\ {x}^{2} - x - 12 + 12 = 0 \\ {x}^{2} - x = 0$

$x(x - 1) = 0 \\ x = 0 \\ x - 1 = 0 \\ x = 1$

s={0,1}

d)

$8 {x}^{2} = 60 + 7 {x}^{2} \\ 8 {x}^{2} - 7 {x}^{2} = 60 \\ {x}^{2} = 60 \\ x = ± \sqrt{60} = ±2 \sqrt{15}$

s={-2√15,2√15}

e)

$8 {x}^{2} = 60 - 7 {x}^{2} \\ 8 {x}^{2} + 7 {x}^{2} = 60 \\ 15 {x}^{2} = 60 \\ {x}^{2} = \frac{60}{15}$

${x}^{2} = 4 \\ x = ± \sqrt{4} = ±2$

s= {-2,2}

f)

${x}^{2} + x(x + 6) = 0 \\ {x}^{2} + {x}^{2} + 6x = 0 \\ 2{x}^{2} + 6x = 0 \\ 2x(x + 3) = 0$

$2x(x + 3) = 0 \\ 2x = 0 \\ x = \frac{0}{2} = 0 \\ x + 3 = 0 \\ x = - 3$

s={-3,0}