Find Positive Values of x in the Quadratic Inequality y = 4x² + 8x

Look at the following function:

$y=4x^2+8x$

Determine for which values of $x$ the following holds:

$f(x) > 0$

To solve this problem, we first need to find the roots of the quadratic function $y = 4x^2 + 8x$.

Let's factor the quadratic equation:

$y = 4x^2 + 8x = 4x(x + 2)$

The roots of this equation are found by setting each factor to zero:

$4x = 0$ gives $x = 0$

$x + 2 = 0$ gives $x = -2$

Thus, the roots are $x = 0$ and $x = -2$. These roots divide the number line into three intervals: $(-\infty, -2)$, $(-2, 0)$, and $(0, \infty)$.

Next, we test a value from each interval to determine where the function is positive:

• For the interval $(-\infty, -2)$, test $x = -3$:
  $y = 4(-3)^2 + 8(-3) = 36 - 24 = 12$ (positive)
• For the interval $(-2, 0)$, test $x = -1$:
  $y = 4(-1)^2 + 8(-1) = 4 - 8 = -4$ (negative)
• For the interval $(0, \infty)$, test $x = 1$:
  $y = 4(1)^2 + 8(1) = 4 + 8 = 12$ (positive)

Based on these tests, $y = 4x^2 + 8x$ is positive in the intervals $x < -2$ and $x > 0$.

Therefore, the solution to the problem is $x > 0$ or $x < -2$.