X2-25≤0

To solve the inequality \(x^2 - 25 \leq 0\), we can first rewrite it in factored form. Notice that \(x^2 - 25\) is a difference of squares: \[ x^2 - 25 = (x - 5)(x + 5) \] So the inequality can be rewritten as: \[ (x - 5)(x + 5) \leq 0 \] Next, we find the critical points by setting the expression equal to zero: \[ (x - 5)(x + 5) = 0 \] This gives us the solutions: \[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \] \[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \] Now we have the critical points \(x = -5\) and \(x = 5\). These points divide the number line into three intervals: 1. \(x < -5\) 2. \(-5 \leq x \leq 5\) 3. \(x > 5\) We will test each interval to see where the product \((x - 5)(x + 5)\) is less than or equal to zero. 1. **Interval 1: \(x < -5\) (e.g., \(x = -6\))** \((x - 5)(x + 5) = (-6 - 5)(-6 + 5) = (-11)(-1) = 11\) (positive) 2. **Interval 2: \(-5 \leq x \leq 5\) (e.g., \(x = 0\))** \((x - 5)(x + 5) = (0 - 5)(0 + 5) = (-5)(5) = -25\) (negative) 3. **Interval 3: \(x > 5\) (e.g., \(x = 6\))** \((x - 5)(x + 5) = (6 - 5)(6 + 5) = (1)(11) = 11\) (positive) From the tests, we see that the product is negative in the interval \([-5, 5]\) and zero at the endpoints. Therefore, the solution to the inequality \(x^2 - 25 \leq 0\) is: \[ \boxed{[-5, 5]} \]