Domain Analysis: Find Valid Inputs for ½x² - 4/9

To find the roots of the quadratic equation $\frac{1}{2}x^2 - \frac{4}{9} = 0$, follow these steps:

• Set the equation to zero: $\frac{1}{2}x^2 - \frac{4}{9} = 0$.
• Multiply the entire equation by 9 to clear the fraction: $9 \times \frac{1}{2}x^2 - 4 = 0$, simplifying to $\frac{9}{2}x^2 = 4$.
• Multiply through by 2 to solve for $x^2$: $9x^2 = 8$.
• Divide both sides by 9: $x^2 = \frac{8}{9}$.
• Take the square root of both sides: $x = \pm \frac{\sqrt{8}}{3}$, simplifying $\sqrt{8}$ to $2\sqrt{2}$.
• Thus the roots are $x = \frac{2\sqrt{2}}{3}$ and $x = -\frac{2\sqrt{2}}{3}$.

These roots divide the number line into intervals: $x < -\frac{2\sqrt{2}}{3}$, $-\frac{2\sqrt{2}}{3} < x < \frac{2\sqrt{2}}{3}$, and $x > \frac{2\sqrt{2}}{3}$.

Evaluate the sign of $y$ in each interval:

• When $x < -\frac{2\sqrt{2}}{3}$, choose a test point and check: $x = -1$, then $y = \frac{1}{2} \times (-1)^2 - \frac{4}{9} = \frac{1}{2} - \frac{4}{9} < 0$.
• When $-\frac{2\sqrt{2}}{3} < x < \frac{2\sqrt{2}}{3}$, choose a test point and check: $x = 0$, then $y = \frac{1}{2} \times 0^2 - \frac{4}{9} = -\frac{4}{9} < 0$.
• When $x > \frac{2\sqrt{2}}{3}$, choose a test point and check: $x = 1$, then $y = \frac{1}{2} \times 1^2 - \frac{4}{9} > 0$.

Therefore, $y$ is positive for $x > \frac{2\sqrt{2}}{3}$ and negative for $-\frac{2\sqrt{2}}{3} < x < \frac{2\sqrt{2}}{3}$, as well as $x < -\frac{2\sqrt{2}}{3}$.

The positive domain for $y$ is $x > \frac{2\sqrt{2}}{3}$, and the negative domain is $x < -\frac{2\sqrt{2}}{3}$ and $-\frac{2\sqrt{2}}{3} < x < \frac{2\sqrt{2}}{3}$.

Thus, the correct answer is:

$x < 0 : -\frac{2\sqrt{2}}{3} < x < \frac{2\sqrt{2}}{3}$

$x > \frac{2\sqrt{2}}{3}$ or $x>0:x<-\frac{2\sqrt{2}}{3}$