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 \( -4 \leq x \leq 1 \).
