Solve (3x − 2)(3x + 2) − 9(x − 1)x = 0.

Step-by-step solution

1. 1

   Expand the first product with the difference-of-squares identity. The pair $(3x-2)(3x+2)$ is of the form $(A-B)(A+B)$:

   $(3x-2)(3x+2) = (3x)^2 - 2^2 = 9x^2 - 4$

   The cross terms $+6x$ and $-6x$ cancel, so there is no linear term from this part.

2. 2

   Expand the second product.

   $9(x-1)x = 9(x^2 - x) = 9x^2 - 9x$

   Distribute the $9$ over both terms; multiplying only the $x^2$ is a common slip.

3. 3

   Subtract and watch the quadratic terms disappear.

   $9x^2 - 4 - (9x^2 - 9x) = 9x^2 - 4 - 9x^2 + 9x = 9x - 4$

   Both expressions carry the same $9x^2$, so the equation is linear, not quadratic — which is why it has only one solution rather than two.

4. 4

   Solve the resulting linear equation.

   $9x - 4 = 0 \ \Longrightarrow \ x = \frac{4}{9}$

5. 5

   Check the root. At $x = \tfrac49$: the first product is $\left(\tfrac{12}{9} - 2\right)\left(\tfrac{12}{9} + 2\right) = \left(-\tfrac23\right)\left(\tfrac{10}{3}\right) = -\tfrac{20}{9}$, and the second is $9\left(\tfrac49 - 1\right)\tfrac49 = 9 \cdot \left(-\tfrac59\right)\cdot\tfrac49 = -\tfrac{20}{9}$. Their difference is $0 \ \checkmark$.