Question

Given that 3 log₃ (2x – 1) = 2 + log₃ (14x – 25)

show that
2x³ – 3x² – 30x + 56 = 0
show that -4 is a root of this cibic equation

Answer 1

Hi,

Explanation:

`3*log_3(2x-1)=2+log_3(14x-25)\\\\log_3((2x-1)^3)-log_3(14x-25)=2\\\\log_3(((2x-1)^3)/(14x-25)) =2\\\\((2x-1)^3)/(14x-25)=9\\\\8x^3-12x^2+6x-1=126x-225\\\\8x^3-12x^2-120x+224=0\\\\P(x)=2x^3-3x^2-30x+56=0\\\\P(-4)=2*(-4)^3-3*(-4)^2-30*(-4)+56\\=-128-48+120+56\\=-176+176\\=0\\`

-4 is a root of P(x)