\( a x ^ { 4 } + b x ^ { 2 } + c = 0 \)

The equation you provided is a quartic equation in the form of \( ax^4 + bx^2 + c = 0 \). To solve this equation, we can make a substitution to simplify it. Let \( y = x^2 \). Then, the equation becomes: \[ a y^2 + b y + c = 0 \] This is a quadratic equation in terms of \( y \). We can solve for \( y \) using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Once we find the values of \( y \), we can substitute back to find \( x \) since \( y = x^2 \). This gives us: \[ x^2 = y \] Thus, the solutions for \( x \) will be: \[ x = \pm \sqrt{y} \] Now, let's summarize the steps: 1. Substitute \( y = x^2 \) into the equation to get \( ay^2 + by + c = 0 \). 2. Use the quadratic formula to solve for \( y \). 3. Substitute back to find \( x \) using \( x = \pm \sqrt{y} \). ### Example If we have a specific example, such as \( 2x^4 + 3x^2 + 1 = 0 \): 1. Substitute \( y = x^2 \): \( 2y^2 + 3y + 1 = 0 \). 2. Use the quadratic formula: \[ y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{-3 \pm \sqrt{9 - 8}}{4} = \frac{-3 \pm 1}{4} \] This gives us: \[ y_1 = \frac{-2}{4} = -\frac{1}{2}, \quad y_2 = \frac{-4}{4} = -1 \] Since both \( y_1 \) and \( y_2 \) are negative, there are no real solutions for \( x \) in this case. The solutions would be complex: \[ x^2 = -\frac{1}{2} \implies x = \pm i \sqrt{\frac{1}{2}} = \pm \frac{i}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} i \] \[ x^2 = -1 \implies x = \pm i \] Thus, the complete set of solutions for this example would be: \[ x = \pm \frac{\sqrt{2}}{2} i, \quad x = \pm i \] If you have specific values for \( a \), \( b \), and \( c \), please provide them, and I can help you solve that specific equation!