√x²+x-2

To solve the inequality \( \sqrt{x^2 + x - 2} < x \), we'll proceed with the following steps: 1. **Isolate the square root**: We need to ensure that both sides are valid and isolate the square root. We start by squaring both sides, but we should keep in mind that when we square both sides, we may introduce extraneous solutions. \[ \sqrt{x^2 + x - 2} < x \] Square both sides (valid only when \( x \geq 0 \)): \[ x^2 + x - 2 < x^2 \] 2. **Simplify the inequality**: Subtract \( x^2 \) from both sides: \[ x - 2 < 0 \] Rearranging gives: \[ x < 2 \] 3. **Consider the domain of the square root**: The expression \( \sqrt{x^2 + x - 2} \) must be defined, so we need to ensure that \( x^2 + x - 2 \geq 0 \). Factor the quadratic: \[ (x - 1)(x + 2) \geq 0 \] The critical points are \( x = 1 \) and \( x = -2 \). We analyze the sign of \( (x - 1)(x + 2) \): - For \( x < -2 \), both factors are negative, hence the product is positive. - For \( -2 < x < 1 \), the factor \( (x - 1) \) is negative and \( (x + 2) \) is positive, hence the product is negative. - For \( x > 1 \), both factors are positive, hence the product is positive. Therefore, \( (x - 1)(x + 2) \geq 0 \) holds for: \[ x \in (-\infty, -2] \cup [1, \infty) \] 4. **Combine the results**: We have two conditions: - \( x < 2 \) - \( x \in (-\infty, -2] \cup [1, \infty) \) Thus, we need the intersection of \( x < 2 \) with \( (-\infty, -2] \cup [1, \infty) \): - From \( x < 2 \) and \( (-\infty, -2] \), we have \( (-\infty, -2] \). - From \( x < 2 \) and \( [1, \infty) \), this region is \( [1, 2) \). 5. **Final solution**: The final solution to the inequality \( \sqrt{x^2 + x - 2} < x \) is: \[ x \in (-\infty, -2] \cup [1, 2) \] Thus, the solution set is \( x \in (-\infty, -2] \cup [1, 2) \).