Questions: Solve: 16 x^2 - 81 = 0 (A) -4/9, 4/9 (B) -81/16, 81/16 (C) -16/81, 16/81 (D) -9/4, 9/4

Transcript text: Solve: $16 x^{2}-81=0$ (A) $-\frac{4}{9}, \frac{4}{9}$ (B) $-\frac{81}{16}, \frac{81}{16}$ (C) $-\frac{16}{81}, \frac{16}{81}$ D $-\frac{9}{4}, \frac{9}{4}$

Solution

Solution Steps

To solve the quadratic equation $16x^2 - 81 = 0$, we can use the following steps:

Recognize that the equation is in the form of $a^2 - b^2 = 0$, which is a difference of squares.
Factor the equation as $(4x - 9)(4x + 9) = 0$.
Set each factor equal to zero and solve for $x$.

Step 1: Recognize the Form of the Equation

The given equation is $16x^2 - 81 = 0$. This is a difference of squares, which can be factored as $a^2 - b^2 = (a - b)(a + b)$.

Step 2: Factor the Equation

We can rewrite $16x^2 - 81$ as $(4x)^2 - 9^2$. Therefore, the equation can be factored as: \[ (4x - 9)(4x + 9) = 0 \]

Step 3: Solve for $x$

Set each factor equal to zero and solve for $x$: \[ 4x - 9 = 0 \quad \text{or} \quad 4x + 9 = 0 \] Solving these equations, we get: \[ 4x - 9 = 0 \implies x = \frac{9}{4} \] \[ 4x + 9 = 0 \implies x = -\frac{9}{4} \]