How do you solve the inequality #x^2 - 4x ≤ -3#?

Question

Eva Adamson · Accepted Answer

# 1 ≤ x ≤ 3 # #x^2−4x≤−3# # x^2 - 4x + 2^2≤ 4 - 3# # (x - 2)^2 ≤ 1# # -1 ≤ x - 2 ≤ 1# # 1 ≤ x ≤ 3 #

Kennedy Adams · Answer

undefined To solve the inequality $ x^2 - 4x \leq -3 $, follow these steps:

1. Move all terms to one side of the inequality:
   $ x^2 - 4x + 3 \leq 0 $

2. Factor the quadratic expression:
   $ (x - 3)(x - 1) \leq 0 $

3. Find the critical points by setting each factor equal to zero and solving for $ x $:
   $ x - 3 = 0 $ and $ x - 1 = 0 $
   $ x = 3 $ and $ x = 1 $

4. Create intervals using the critical points:
   The critical points divide the number line into three intervals: $ (-\infty, 1) $, $ (1, 3) $, and $ (3, +\infty) $.

5. Test a value from each interval in the inequality to determine the solution set:
   - For $ x = 0 $ (in $ (-\infty, 1) $): $ (0 - 3)(0 - 1) = (-3)(-1) = 3 $, which is greater than 0. So, this interval is not part of the solution.
   - For $ x = 2 $ (in $ (1, 3) $): $ (2 - 3)(2 - 1) = (-1)(1) = -1 $, which is less than or equal to 0. So, this interval is part of the solution.
   - For $ x = 4 $ (in $ (3, +\infty) $): $ (4 - 3)(4 - 1) = (1)(3) = 3 $, which is greater than 0. So, this interval is not part of the solution.

Therefore, the solution to the inequality is $ x \in (1, 3] $.