Solve the Quadratic Inequality: When is y = x² + 4x Positive?

Look at the following function:

$y=x^2+4x$

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

$f(x) > 0$

To determine when the function $f(x) = x^2 + 4x$ is positive, we start by analyzing the quadratic expression. The expression can be factored as:

$x^2 + 4x = x(x + 4)$

To find when this is greater than zero, identify the roots of the equation $x(x + 4) = 0$. Solving this, we find the roots to be:

• $x = 0$
• $x = -4$

These roots split the real number line into three intervals, which we must analyze to determine where the function is positive:

• Interval 1: $x < -4$
• Interval 2: $-4 < x < 0$
• Interval 3: $x > 0$

We test a point from each interval to determine the sign of the function:

• For $x < -4$: Choose $x = -5$. Then $f(-5) = (-5)((-5) + 4) = 5$, which is positive.
• For $-4 < x < 0$: Choose $x = -2$. Then $f(-2) = (-2)((-2) + 4) = -4$, which is negative.
• For $x > 0$: Choose $x = 1$. Then $f(1) = (1)((1) + 4) = 5$, which is positive.

From this analysis, the function $f(x)$ is positive in the intervals:

$x < -4$ and $x > 0$.

Therefore, the correct choice is:

$x < -4$ or $x > 0$