√x²+x-2

To solve the inequality \( \sqrt{x^2 + x - 2} < x \), we start by isolating the square root: 1. **Square both sides** (keeping in mind that squaring can introduce extraneous solutions, so we will verify later): \[ x^2 + x - 2 < x^2 \] 2. **Subtract \( x^2 \) from both sides**: \[ x - 2 < 0 \] 3. **Rearranging gives**: \[ x < 2 \] Next, we need to ensure our solution doesn't include extraneous solutions introduced by squaring. The original inequality has a square root, so we also need to ensure that the expression under the square root is non-negative: 4. **Set the expression under the square root greater than or equal to zero**: \[ x^2 + x - 2 \geq 0 \] 5. **Factor the quadratic**: \[ (x - 1)(x + 2) \geq 0 \] 6. **Find the critical points**: The roots of the equation are \( x = 1 \) and \( x = -2 \). 7. **Test intervals** around the critical points: - For \( x < -2 \): Choose \( x = -3 \), then \( (-3 - 1)(-3 + 2) = (-4)(-1) = 4 \geq 0 \) (true). - For \( -2 < x < 1 \): Choose \( x = 0 \), then \( (0 - 1)(0 + 2) = (-1)(2) = -2 < 0 \) (false). - For \( x > 1 \): Choose \( x = 2 \), then \( (2 - 1)(2 + 2) = (1)(4) = 4 \geq 0 \) (true). 8. **Determine the solution intervals** from the checks: - The inequality is satisfied in the intervals \( (-\infty, -2] \) and \( [1, \infty) \). 9. **Combine the results**: We need \( x < 2 \) from step 3 and \( (x \leq -2) \cup (x \geq 1) \) from step 8. - The solutions \( x < 2 \) and \( x \geq 1 \) gives \( 1 \leq x < 2 \). - The set \( (-\infty, -2] \) does not intersect with \( x < 2 \). 10. **Final Consideration**: Therefore, the solution to the inequality \( \sqrt{x^2 + x - 2} < x \) is: \[ \boxed{[1, 2)} \]