Solve x\(^2\) + 3x - 4 ≤ 0

This question asks you to solve a quadratic inequality: \( x^2 + 3x - 4 \leq 0 \)

To solve this, follow these steps:

• Step 1: Rewrite the inequality as an equation to find the boundary points (known as the "roots" or "zeros"):
  Set \( x^2 + 3x - 4 = 0 \).


• Step 2: Factor the quadratic, if possible:
  \( x^2 + 3x - 4 = (x+4)(x-1) = 0 \)
  So, the roots are \( x = -4 \) and \( x = 1 \).


• Step 3: Plot these roots on a number line. They divide the real number line into three intervals:
  • \( x < -4 \)
  • \( -4 \leq x \leq 1 \)
  • \( x > 1 \)


• Step 4: Determine which interval(s) satisfy the inequality. Pick a test value from each interval:
  • For \( x < -4 \), try \( x = -5 \):
    \[ (-5)^2 + 3(-5) - 4 = 25 - 15 - 4 = 6 \]
    \( 6 \) is not less than or equal to 0.
  • For \( -4 \leq x \leq 1 \), try \( x = 0 \):
    \[ (0)^2 + 3(0) - 4 = -4 \]
    \(-4 \leq 0\) is true.
  • For \( x > 1 \), try \( x = 2 \):
    \[ (2)^2 + 3(2) - 4 = 4 + 6 - 4 = 6 \]
    Again, \( 6 \) is not less than or equal to 0.


• Step 5: Don’t forget to check the endpoints:
  • At \( x = -4 \):
    \[ (-4)^2 + 3(-4) - 4 = 16 - 12 - 4 = 0 \] \(\ 0 \leq 0\) is true.
  • At \( x = 1 \):
    \[ (1)^2 + 3(1) - 4 = 1 + 3 - 4 = 0 \] \(\ 0 \leq 0\) is true.


• Conclusion: The solution is the interval \( -4 \leq x \leq 1 \).

Why? The quadratic expression is less than or equal to zero exactly between its roots (including the endpoints). Outside this interval, the expression is positive.

Graphical intuition: The graph of \( y = x^2 + 3x - 4 \) is a parabola opening upwards. It is below or on the x-axis between the roots, i.e., for all \( x \) values between \( -4 \) and \( 1 \), including the endpoints.

So, the correct solution is all \( x \) such that \( -4 \leq x \leq 1 \).