√x²+x-2

To solve the inequality \( \sqrt{x^2 + x - 2} < c \), we need to consider when the expression under the square root is non-negative since the square root of a negative number is not defined in the set of real numbers. **Step 1: Determine when the expression under the square root is non-negative.** We set up the inequality: \[ x^2 + x - 2 \geq 0 \] To solve this quadratic inequality, we first find the roots of the equation: \[ x^2 + x - 2 = 0 \] Using the quadratic formula, where \( a = 1, b = 1, c = -2 \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} \] Calculating the roots: \[ x = \frac{2}{2} = 1 \quad \text{and} \quad x = \frac{-4}{2} = -2 \] Thus, the roots are \( x = -2 \) and \( x = 1 \). **Step 2: Analyze the intervals given by the roots.** Now we test the intervals \( (-\infty, -2) \), \( (-2, 1) \), and \( (1, \infty) \) to see where \( x^2 + x - 2 \geq 0 \): - For the interval \( (-\infty, -2) \), pick \( x = -3 \): \[ (-3)^2 + (-3) - 2 = 9 - 3 - 2 = 4 \quad (\text{which is } \geq 0) \] - For the interval \( (-2, 1) \), pick \( x = 0 \): \[ 0^2 + 0 - 2 = -2 \quad (\text{which is } < 0) \] - For the interval \( (1, \infty) \), pick \( x = 2 \): \[ 2^2 + 2 - 2 = 4 + 2 - 2 = 4 \quad (\text{which is } \geq 0) \] Thus, \( x^2 + x - 2 \geq 0 \) in the intervals \( (-\infty, -2] \) and \( [1, \infty) \). **Step 3: Combine with the original inequality.** Now we revisit the original inequality \( \sqrt{x^2 + x - 2} < c \). We must have \( x^2 + x - 2 < c^2 \) where it is valid. Setting up the combined inequalities: 1. \( x^2 + x - 2 < c^2 \) 2. \( x^2 + x - 2 \geq 0 \) implies \( x \in (-\infty, -2] \cup [1, \infty) \) **Step 4: Solve the inequality \( x^2 + x - (2 + c^2) < 0 \).** We can again find the roots of \( x^2 + x - (2 + c^2) = 0 \): Using the quadratic formula: \[ x = \frac{-1 \pm \sqrt{1 + 4(2 + c^2)}}{2} \] This gives the roots and we can analyze the intervals created by these roots. The solutions will be where the parabola is below the x-axis. **Step 5: Combine results.** Finally, we will take the intersections of those intervals where the root conditions are met. You will need to find the exact range for \( c \) to finalize the solution. From here: - For \( c > 0 \), analyze what happens with the discriminants and the roots you'll find through quadratic analysis. In conclusion, the solution will be an intersection of the areas where both conditions are satisfied. Let me know if you need help analyzing the discriminant or finding specific values for \( c \)!