√x -20≤x-2

To solve the inequality \(\sqrt{x} - 20 \leq x - 2\), we will first isolate the square root term. 1. Start by rearranging the inequality: \[ \sqrt{x} \leq x - 2 + 20 \] Simplifying this gives: \[ \sqrt{x} \leq x + 18 \] 2. Next, we will square both sides to eliminate the square root. Remember that when we square both sides, we need to be cautious about the domains and possible extraneous solutions: \[ (\sqrt{x})^2 \leq (x + 18)^2 \] This simplifies to: \[ x \leq x^2 + 36x + 324 \] Rearranging this gives: \[ 0 \leq x^2 + 35x + 324 \] or \[ x^2 + 35x + 324 \geq 0 \] 3. Now, we can find the roots of the quadratic equation \(x^2 + 35x + 324 = 0\) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a = 1\), \(b = 35\), and \(c = 324\): \[ b^2 - 4ac = 35^2 - 4 \cdot 1 \cdot 324 = 1225 - 1296 = -71 \] Since the discriminant is negative \((-71)\), this quadratic does not have real roots. This means \(x^2 + 35x + 324\) is always positive (because the parabola opens upwards), which implies: \[ x^2 + 35x + 324 \geq 0 \text{ for all } x. \] 4. However, we also need to consider the domain of the original inequality. The term \(\sqrt{x}\) requires \(x \geq 0\). 5. Thus, combining these observations, our solution is: \[ x \geq 0 \] In conclusion, the solution to the inequality \(\sqrt{x} - 20 \leq x - 2\) is: \[ x \geq 0 \]