Determine Positive Values of x for the Quadratic Function y = x² + 5x + 4

To determine where the function $y = x^2 + 5x + 4$ is greater than zero, we first find its roots by setting $x^2 + 5x + 4 = 0$.

Step 1: Factor the quadratic equation.

The expression $x^2 + 5x + 4$ can be factored as $(x + 1)(x + 4) = 0$.

Step 2: Solve for the roots.

Setting each factor to zero gives the roots as follows:
$x + 1 = 0 \Rightarrow x = -1$
$x + 4 = 0 \Rightarrow x = -4$

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

• Interval 1: $x < -4$. Pick $x = -5$, then $(x + 1)(x + 4) = (-5 + 1)(-5 + 4) = (-4)(-1) = 4 > 0.$
• Interval 2: $-4 < x < -1$. Pick $x = -3$, then $(x + 1)(x + 4) = (-3 + 1)(-3 + 4) = (-2)(1) = -2 < 0.$
• Interval 3: $x > -1$. Pick $x = 0$, then $(x + 1)(x + 4) = (0 + 1)(0 + 4) = 1 \times 4 = 4 > 0.$

Conclusion: The function $y = x^2 + 5x + 4$ is positive when $x < -4$ or $x > -1$.

Thus, the solution is $x > -1$ or $x < -4$.