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

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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! The solutions to the equation $$5x^2 + 2x + k = 0$$ are: $$ x = \frac{ -2 \pm \sqrt{4 - 2k} }{5} $$ These solutions are real if $$ k \leq 2 $$.

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Solución

1. Standard Quadratic Formula

2. Discriminant Analysis

3. Conditions for Real Solutions

4. Graphical Interpretation

5. Example Calculations

6. Summary

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