Which equation is equivalent to
[tex]\[ f(x) = 16x^4 - 81 = 0? \][/tex]

A. (4x^2 - 9)(4x^2 + 9) = 0

B. (2x - 3)(2x + 3)(4x^2 + 9) = 0

C. (4x^2 - 81)(x^2 - 1) = 0

D. (4x^2 - 9)(4x^2 - 9) = 0



Answer :

To solve the equation [tex]\( f(x) = 0 \)[/tex], where [tex]\( f(x) = 16x^4 - 81 \)[/tex], we will follow these steps:

1. Start with the given equation:
[tex]\[ 16x^4 - 81 = 0 \][/tex]

2. Add 81 to both sides of the equation to isolate the quartic term:
[tex]\[ 16x^4 = 81 \][/tex]

3. Divide both sides by 16 to solve for [tex]\( x^4 \)[/tex]:
[tex]\[ x^4 = \frac{81}{16} \][/tex]

4. Take the fourth root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \pm \sqrt[4]{\frac{81}{16}} \][/tex]

5. Simplify the expression under the fourth root:
[tex]\[ x = \pm \sqrt[4]{\left(\frac{9}{4}\right)^2} \][/tex]

6. Since [tex]\(\sqrt[4]{a^2} = \sqrt{a}\)[/tex], we get:
[tex]\[ x = \pm \sqrt{\frac{9}{4}} = \pm \frac{3}{2} \][/tex]

Thus, we have two real solutions:
[tex]\[ x = \frac{3}{2} \quad \text{and} \quad x = -\frac{3}{2} \][/tex]

7. Consider the complex roots by recognizing that taking the fourth root in the complex plane also gives complex solutions:
[tex]\[ x = \pm \frac{3i}{2} \][/tex]

So, the complete solution set for the equation is:
[tex]\[ x = \frac{3}{2}, \quad x = -\frac{3}{2}, \quad x = \frac{3i}{2}, \quad x = -\frac{3i}{2} \][/tex]

Hence, the final solutions to the equation [tex]\( 16x^4 - 81 = 0 \)[/tex] are:
[tex]\[ -\frac{3}{2}, \frac{3}{2}, -\frac{3i}{2}, \frac{3i}{2} \][/tex]