Question

Solve the quadratic by factoring.
x^4 + 4x^4 - 45

Answer 1

It seems like there's a small mistake in the expression you provided. The expression \( x^4 + 4x^4 - 45 \) can be simplified before proceeding. Combining like terms gives:

\[
5x^4 - 45
\]

To solve the equation \( 5x^4 - 45 = 0 \) by factoring, we can first factor out the common factor, which is 5:

\[
5(x^4 - 9) = 0
\]

Next, we recognize that \( x^4 - 9 \) can be factored as a difference of squares:

\[
x^4 - 9 = (x^2 - 3)(x^2 + 3)
\]

Now we can rewrite the equation as:

\[
5(x^2 - 3)(x^2 + 3) = 0
\]

Setting each factor equal to zero gives us:

1. \( x^2 - 3 = 0 \)
2. \( x^2 + 3 = 0 \)

Solving \( x^2 - 3 = 0 \):

\[
x^2 = 3 \quad \Rightarrow \quad x = \pm \sqrt{3}
\]

Solving \( x^2 + 3 = 0 \):

\[
x^2 = -3 \quad \Rightarrow \quad x = \pm i\sqrt{3}
\]

Thus, the complete set of solutions to the original equation \( 5x^4 - 45 = 0 \) is:

\[
x = \sqrt{3}, \quad x = -\sqrt{3}, \quad x = i\sqrt{3}, \quad x = -i\sqrt{3}
\]