5x^4-25x^2-180=0
To solve the equation \(5x^4 - 25x^2 - 180 = 0\), we can start by making a substitution to simplify the equation. Let \(y = x^2\). Then the equation becomes:
\[ 5y^2 - 25y - 180 = 0 \]
Next, we can simplify this equation by dividing through by 5:
\[ y^2 - 5y - 36 = 0 \]
Now we can use the quadratic formula to solve for \(y\):
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = -5\), and \(c = -36\). Substituting in these values:
\[ y = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-36)}}{2 \cdot 1} \] \[ y = \frac{5 \pm \sqrt{25 + 144}}{2} \] \[ y = \frac{5 \pm \sqrt{169}}{2} \] \[ y = \frac{5 \pm 13}{2} \]
Now, calculating the two possible values for \(y\):
- \(y = \frac{18}{2} = 9\)
- \(y = \frac{-8}{2} = -4\)
Since \(y = x^2\), we only consider the non-negative value, \(y = 9\). Thus:
\[ x^2 = 9 \]
Taking the square root of both sides gives:
\[ x = 3 \quad \text{or} \quad x = -3 \]
The negative value \(y = -4\) does not yield real solutions for \(x\) since \(x^2\) cannot be negative.
Thus, the final solutions to the original equation \(5x^4 - 25x^2 - 180 = 0\) are:
\[ \boxed{3 \text{ and } -3} \]
