Question

can anyone solve any of the 4 questions

Answer 1

x(x+1)(x+2)(x+3) = 120

or (x^2+3x)(x^2+3x+2) = 120

Let x^+3x = y

Then equation becomes y(y+2) = 120

or y^2+2y = 120 or (y+12)(y-10) = 0

Thus y = -12 or y = 10

or x^2+3x = -12 or x^2+3x = 10

The first equation dies not have real roots

Solving the second equation we have

(x+5)(x-2) = 0 or x = -5 or 2

Answer 2

4. here f (x)= 3x^2-2x+1
g (x)=2x^2-"2" 《question is incomplete here, so i imagined 2》
now
3f (x)=9x^2-6x+3
4g (x)=8x^2-8
as 3f (x)=4g (x)
=>9x^2-6x+3=8x^2-8
=>x^2-6x+5=0
=>(x+2)(x+4)=0
so x=-2 or x=-4