MATH SOLVE

2 months ago

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

Accepted Solution

A:

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

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