√x²+x-2<x
To solve the inequality \( \sqrt{x^2 + x - 2} < x \), we start by isolating the square root:
Square both sides (keeping in mind that squaring can introduce extraneous solutions, so we will verify later): \[ x^2 + x - 2 < x^2 \]
Subtract \( x^2 \) from both sides: \[ x - 2 < 0 \]
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:
Set the expression under the square root greater than or equal to zero: \[ x^2 + x - 2 \geq 0 \]
Factor the quadratic: \[ (x - 1)(x + 2) \geq 0 \]
Find the critical points: The roots of the equation are \( x = 1 \) and \( x = -2 \).
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).
Determine the solution intervals from the checks:
- The inequality is satisfied in the intervals \( (-\infty, -2] \) and \( [1, \infty) \).
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 \( x < 2 \).
Final Consideration: Therefore, the solution to the inequality \( \sqrt{x^2 + x - 2} < x \) is: \[ \boxed{[1, 2)} \]
