For what value of xis the square of the binomial 3x+1 is 9 times greater than the square of the binomial x–2?

X = 5/6.

The square of 3x+1 is written as (3x+1)².  The square of x-2 is (x-2)².  Using the information given to us, we want to solve the equation

(3x+1)²=9(x-2)²
(3x+1)(3x+1)=9(x-2)(x-2)

Multiplying the first two binomials, we have:
3x*3x + 1*3x + 1*3x + 1*1 = 9(x-2)(x-2)
9x²+3x+3x+1 = 9(x-2)(x-2)
9x²+6x+1 = 9(x-2)(x-2)

Multiplying the second two binomials, we have:
9x²+6x+1 = 9(x*x-2*x-2*x-2(-2))
9x²+6x+1 = 9(x²-2x-2x+4)
9x²+6x+1 = 9(x²-4x+4)

Using the distributive property gives us
9x²+6x+1 = 9*x²-9*4x+9*4
9x²+6x+1 = 9x²-36x+36

Subtracting 9x² from both sides leaves us
6x+1 = -36x + 36

Adding 36x to both sides we get
42x+1 = 36

Subtracting 1 from both sides we have
42x = 35

Divide both sides by 42:
42x/42 = 35/42
x = 35/42 = 5/6