can anyone help me with this? also, take note that the word “ hence ” requires the use of the previous solution used in the earlier parts.

can anyone help me with this also take note that the word hence requires the use of the previous solution used in the earlier parts class=

Respuesta :

(a) x = +/- 3 (or 0)

-Because x is being squared, it does not matter if it is positive or negative, as when any number is squared the product is always positive.

3x^2 - 9x = 0

3x(x - 3) = 0

x = 0

x = +/- 3

(b) x = 6, -4

x^2 - 2x - 24 = 0

(x - 6)(x + 4) = 0

x = 6

x = -4

(c) y = 8, -2

(y - 2)^2 - 2(y - 2) - 24 = 0

(y^2 - 4y + 4) - 2y + 4 - 24 = 0

y^2 - 6y - 16 = 0

(y - 8)(y + 2) = 0

y = 8

y = -2

Hope this helps! :)