Question

Find the real or imaginary solutions of the equation by factoring.
x^4 -50x^2=-49
The solutions are____. (use a comma to separate answers as needed)

Answer 1

To solve the equation \( x^4 - 50x^2 = -49 \), first, we can rearrange it into a standard form:

\[
x^4 - 50x^2 + 49 = 0
\]

Next, we can make a substitution to simplify the expression. Let \( y = x^2 \). Then the equation becomes:

\[
y^2 - 50y + 49 = 0
\]

This is a quadratic equation which we can solve using the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = -50 \), and \( c = 49 \). Plugging these values into the formula gives:

\[
y = \frac{50 \pm \sqrt{(-50)^2 - 4 \cdot 1 \cdot 49}}{2 \cdot 1}
\]
\[
y = \frac{50 \pm \sqrt{2500 - 196}}{2}
\]
\[
y = \frac{50 \pm \sqrt{2304}}{2}
\]
\[
y = \frac{50 \pm 48}{2}
\]

Now, we compute the two potential values for \( y \):

1.
\[
y = \frac{50 + 48}{2} = \frac{98}{2} = 49
\]
2.
\[
y = \frac{50 - 48}{2} = \frac{2}{2} = 1
\]

Now we revert back to \( x \) by substituting \( y \) back to \( x^2 \):

1. For \( y = 49 \):
\[
x^2 = 49 \implies x = \pm 7
\]

2. For \( y = 1 \):
\[
x^2 = 1 \implies x = \pm 1
\]

Thus, the solutions to the original equation are:

\[
x = 7, -7, 1, -1
\]

The complete solution set is:

\[
\boxed{7, -7, 1, -1}
\]