Finding X: Solve y = x² + 5x + 4 for Negative Output

Question

Look at the following function:

y=x2+5x+4 y=x^2+5x+4

Determine for which values of x x the following is true:

f(x) < 0

Step-by-Step Solution

To determine where the function f(x)=x2+5x+4 f(x) = x^2 + 5x + 4 is less than zero, we will first factor the quadratic expression.

Step 1: Factor the quadratic function.

  • The quadratic x2+5x+4 x^2 + 5x + 4 can be factored into (x+4)(x+1) (x + 4)(x + 1) .

Step 2: Find the roots of the quadratic equation.

  • Set each factor equal to zero: x+4=0 x + 4 = 0 and x+1=0 x + 1 = 0 .
  • The solutions are x=4 x = -4 and x=1 x = -1 .

Step 3: Determine the sign of the quadratic in the intervals defined by these roots.

  • Consider the intervals (,4) (-\infty, -4) , (4,1) (-4, -1) , and (1,) (-1, \infty) .
  • Pick test points from each interval: for (,4) (-\infty, -4) , choose x=5 x = -5 ; for (4,1) (-4, -1) , choose x=2 x = -2 ; for (1,) (-1, \infty) , choose x=0 x = 0 .
  • Evaluate the sign of (x+4)(x+1) (x+4)(x+1) at each test point:
    • At x=5 x = -5 , (x+4)(x+1)=(1)(4)=4 (x+4)(x+1) = (-1)(-4) = 4 (positive).
    • At x=2 x = -2 , (x+4)(x+1)=(2)(1)=2 (x+4)(x+1) = (2)(-1) = -2 (negative).
    • At x=0 x = 0 , (x+4)(x+1)=(4)(1)=4 (x+4)(x+1) = (4)(1) = 4 (positive).
  • Therefore, the function is negative in the interval (4,1) (-4, -1) .

Consequently, the solution is 4<x<1 -4 < x < -1 .

The correct choice from the options given is choice 4.

Answer

-4 < x < -1