Question

solve fast plz...........​

Answer 1

Step-by-step explanation:

let x² + 5x = y

(y + 4)(y + 6) = 120

=> y² + 10y + 24 = 120

=> y² + 10y - 96 = 0

=> y² + 16y - 6y - 96 = 0

=> y(y + 16) - 6(y + 16) = 0

=> (y - 6)(y + 16) = 0

=> y - 6 = 0 => y = 6.................1

=> y + 16 = 0 => y = -16...........2

from equation 1,

x² + 5x = 6

=> x² + 5x - 6 = 0

=> x² + 6x - x - 6 = 0

=> x(x + 6) -1(x + 6) = 0

=> (x - 1)(x + 6) = 0

=> x - 1 = 0 => x = 1

=> x + 6 = 0 => x = -6

from equation 2,

x² + 5x = -16

=> x² + 5x + 16 = 0

a = 1, b = 5 and c = 16

x = [-5 ± √5² - 4*1*16] / 2*1

x = [-5 ± √-39]/2

as D < 0, no real solution at y = -16

therefore solutions are x = 1 and x = -6