Factor 4x^2 - 9, then solve 4x^2 - 9 = 0.

Step-by-step solution

1. 1

   Check that both terms are perfect squares. This is the precondition for the difference-of-squares pattern:

   $4x^{2}=(2x)^{2},\qquad9=3^{2}$

   The coefficient $4$ is itself a square, so the square root of $4x^{2}$ is $2x$, not $4x$ — halving the coefficient instead of taking its root is the usual mistake.

2. 2

   Apply the identity. With $a^{2}-b^{2}=(a-b)(a+b)$ and $a=2x$, $b=3$:

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

   There is no middle term to account for, because the cross terms $-6x$ and $+6x$ cancel — which is exactly why this pattern looks like a two-term expression.

3. 3

   Verify by expanding. $(2x-3)(2x+3)=4x^{2}+6x-6x-9=4x^{2}-9$ ✓, confirmed at $40$ integer values ✓. Note that $4x^{2}+9$, the sum of squares, does not factor over the reals — the minus sign is essential.

4. 4

   Solve the equation with the zero-product property.

   $(2x-3)(2x+3)=0\quad\Longrightarrow\quad2x-3=0\ \text{ or }\ 2x+3=0$

   $x=\frac{3}{2}\qquad\text{or}\qquad x=-\frac{3}{2}$

5. 5

   Check both roots. At $x=\tfrac32$: $4\cdot\tfrac94-9=9-9=0$ ✓. At $x=-\tfrac32$: $4\cdot\tfrac94-9=0$ ✓ (squaring removes the sign). The same answers follow from the square-root route: $4x^{2}=9\Rightarrow x^{2}=\tfrac94\Rightarrow x=\pm\tfrac32$ ✓.