Question

The solution set of the inequality \( \sqrt{x^2 + x - 2}>(1 - x) \) is:

• A. \((- \infty, 2)\)
• B. \((- \infty, -2)\)
• C. \( (1, \infty) \)
• D. \( (0, \infty) \)

Hint:

Always consider the domain of all functions involved in an inequality. For square roots, ensure the expression under the root is non-negative. Analyze the sign of each part of the inequality to find the valid range.

Answer 1

The Correct Option is C

To find the solution set for the inequality \( \sqrt{x^2 + x - 2}>(1 - x) \), we first consider the domain of the square root function, which requires: \[ x^2 + x - 2 \geq 0 \] Factoring the quadratic, we get: \[ (x - 1)(x + 2) \geq 0 \] The solution to this inequality is \( x \leq -2 \) or \( x \geq 1 \). Next, since the square root function is always non-negative, \( \sqrt{x^2 + x - 2} \geq 0 \), the right side \( (1 - x) \) must also be non-negative for the inequality to hold. Thus: \[ 1 - x>0 \Rightarrow x<1 \] However, to satisfy \( \sqrt{x^2 + x - 2}>(1 - x) \), where \( (1 - x) \) is also non-negative, we focus only on the intersection of \( x \geq 1 \) and analyze further. For \( x \geq 1 \), \( (1 - x) \) becomes non-positive, hence the square root being non-negative is always greater. Therefore, the values that satisfy both the original inequality and the domain constraints are \( x>1 \), so the solution set is: \[ x \in (1, \infty) \]