What are the solutions of [tex]$4x^4 - 4x^2 = 8?$[/tex]

A. [tex]$x = \pm i$[/tex]

B. [tex][tex]$x = \pm 1$[/tex][/tex]

C. [tex]$x = \pm \sqrt{2}$[/tex]

D. [tex]$x = \pm 2$[/tex]



Answer :

Let's solve the given equation [tex]\(4 x^4 - 4 x^2 = 8\)[/tex] step by step.

1. Rewrite the equation:
Start by moving all terms to one side of the equation:
[tex]\[ 4 x^4 - 4 x^2 - 8 = 0 \][/tex]

2. Substitute [tex]\(\displaystyle u = x^2\)[/tex]:
Let [tex]\(u = x^2\)[/tex]. This substitution simplifies the equation because [tex]\(x^4\)[/tex] becomes [tex]\(u^2\)[/tex]. The equation then transforms into:
[tex]\[ 4u^2 - 4u - 8 = 0 \][/tex]

3. Solve the quadratic equation:
We now solve for [tex]\(u\)[/tex] by using the quadratic formula [tex]\(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\( a = 4 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = -8 \)[/tex].
[tex]\[ u = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-8)}}{2 \cdot 4} \][/tex]
[tex]\[ u = \frac{4 \pm \sqrt{16 + 128}}{8} \][/tex]
[tex]\[ u = \frac{4 \pm \sqrt{144}}{8} \][/tex]
[tex]\[ u = \frac{4 \pm 12}{8} \][/tex]

This results in two possible values for [tex]\(u\)[/tex]:
[tex]\[ u = \frac{4 + 12}{8} = \frac{16}{8} = 2 \][/tex]
[tex]\[ u = \frac{4 - 12}{8} = \frac{-8}{8} = -1 \][/tex]

4. Back-substitute [tex]\( \displaystyle u = x^2 \)[/tex]:

- For [tex]\(u = 2\)[/tex]:
[tex]\[ x^2 = 2 \][/tex]
Taking the square root of both sides:
[tex]\[ x = \pm \sqrt{2} \][/tex]

- For [tex]\(u = -1\)[/tex]:
[tex]\[ x^2 = -1 \][/tex]
Taking the square root of both sides:
[tex]\[ x = \pm i \][/tex]

5. Collect all solutions:
The solutions are:
[tex]\[ x = \pm \sqrt{2} \quad \text{and} \quad x = \pm i \][/tex]

6. Compare with given choices:
Among the provided options:
- [tex]\(\pm i\)[/tex] matches.
- [tex]\(\pm \sqrt{2}\)[/tex] matches.

Thus, the correct solutions for the equation [tex]\(4 x^4 - 4 x^2 = 8\)[/tex] are indeed
[tex]\[ x = \pm \sqrt{2} \][/tex]
and
[tex]\[ x = \pm i. \][/tex]