\( x = \frac { - 2 \pm \sqrt { 4 - 2 k } } { 5 } \)

The expression you've provided is: \[ x = \frac{ -2 \pm \sqrt{4 - 2k} }{5} \] This appears to be the solution to a quadratic equation of the form: \[ 5x^2 + 2x + k = 0 \] Let's break down and analyze this expression step by step. ### 1. **Standard Quadratic Formula** The general form of a quadratic equation is: \[ ax^2 + bx + c = 0 \] The solutions (roots) to this equation are given by the quadratic formula: \[ x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{2a} \] Comparing this with your expression: \[ x = \frac{ -2 \pm \sqrt{4 - 2k} }{5} \] We can identify the coefficients: - \( a = 5 \) - \( b = 2 \) - \( c = k \) ### 2. **Discriminant Analysis** The discriminant \( D \) of a quadratic equation is: \[ D = b^2 - 4ac \] In your solution, the expression under the square root is \(4 - 2k\), which corresponds to the discriminant: \[ D = 4 - 2k \] The discriminant determines the nature of the roots: - **If \( D > 0 \)**: There are two distinct real roots. - **If \( D = 0 \)**: There is exactly one real root (a repeated root). - **If \( D < 0 \)**: There are two complex conjugate roots. ### 3. **Conditions for Real Solutions** To ensure the roots are real, the discriminant must be non-negative: \[ 4 - 2k \geq 0 \] Solving for \( k \): \[ 4 - 2k \geq 0 \\ -2k \geq -4 \\ k \leq 2 \] **Conclusion:** For the quadratic equation \(5x^2 + 2x + k = 0\), real solutions for \( x \) exist if and only if \( k \leq 2 \). ### 4. **Graphical Interpretation** If you were to graph the quadratic equation \(5x^2 + 2x + k = 0\): - The parabola opens upwards since the coefficient of \( x^2 \) (which is 5) is positive. - The vertex of the parabola shifts vertically based on the value of \( k \). - When \( k \leq 2 \), the parabola intersects the x-axis at two points (for \( k < 2 \)) or just touches the x-axis at one point (for \( k = 2 \)). - For \( k > 2 \), the parabola does not intersect the x-axis, implying no real roots. ### 5. **Example Calculations** Let's calculate the roots for specific values of \( k \): **a. When \( k = 0 \):** \[ x = \frac{ -2 \pm \sqrt{4 - 0} }{5} = \frac{ -2 \pm 2 }{5} \] This gives two solutions: \[ x = \frac{ -2 + 2 }{5} = 0 \\ x = \frac{ -2 - 2 }{5} = -\frac{4}{5} \] **b. When \( k = 2 \):** \[ x = \frac{ -2 \pm \sqrt{4 - 4} }{5} = \frac{ -2 \pm 0 }{5} = -\frac{2}{5} \] There is exactly one real root: \( x = -\frac{2}{5} \). **c. When \( k = 3 \):** \[ x = \frac{ -2 \pm \sqrt{4 - 6} }{5} = \frac{ -2 \pm \sqrt{-2} }{5} = \frac{ -2 \pm i\sqrt{2} }{5} \] Here, the solutions are complex numbers since the discriminant is negative. ### 6. **Summary** - The given expression represents the solutions to the quadratic equation \(5x^2 + 2x + k = 0\). - Real solutions exist only when \( k \leq 2 \). - The discriminant \( D = 4 - 2k \) determines the nature of the roots: - Two distinct real roots if \( k < 2 \). - One real repeated root if \( k = 2 \). - Two complex conjugate roots if \( k > 2 \). If you have a specific question or need further clarification on a particular aspect of this expression, feel free to ask!