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 \( y \):
\[ y = \frac{50 + 48}{2} = \frac{98}{2} = 49 \] 2. \[ y = \frac{50 - 48}{2} = \frac{2}{2} = 1 \]
Now we revert back to \( x \) by substituting \( y \) back to \( x^2 \):
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For \( y = 49 \): \[ x^2 = 49 \implies x = \pm 7 \]
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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:
\[ \boxed{7, -7, 1, -1} \]