Questions: Find all real solutions of the equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.) 2 x^4 + 8 x^2 + 5 = 0

Solution Steps

To solve the equation \(2x^4 + 8x^2 + 5 = 0\), we can use a substitution method. Let \(y = x^2\), which transforms the equation into a quadratic equation in terms of \(y\): \(2y^2 + 8y + 5 = 0\). We can then solve this quadratic equation using the quadratic formula. Once we find the values of \(y\), we substitute back to find the corresponding values of \(x\).

We start with the equation \(2x^4 + 8x^2 + 5 = 0\). To simplify, we use the substitution \(y = x^2\), transforming the equation into a quadratic form: \[2y^2 + 8y + 5 = 0.\]

We solve the quadratic equation \(2y^2 + 8y + 5 = 0\) using the quadratic formula: \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\] where \(a = 2\), \(b = 8\), and \(c = 5\).

The discriminant is: \[b^2 - 4ac = 8^2 - 4 \times 2 \times 5 = 64 - 40 = 24.\]

The solutions for \(y\) are: \[y_1 = \frac{-8 + \sqrt{24}}{4} = -0.7753,\] \[y_2 = \frac{-8 - \sqrt{24}}{4} = -3.2247.\]

Since \(y = x^2\), we solve for \(x\) by taking the square root of \(y_1\) and \(y_2\). However, both \(y_1\) and \(y_2\) are negative, which means they do not yield real solutions for \(x\) because the square root of a negative number is not real.

Final Answer