1. **State the problem:** Calculate the area of a square given length $= \frac{x-1}{x+1}$ and width $= \frac{2x}{15}$. 2. **Recall the formula for the area of a square:** $$\text{Area} = \text{side} \times \text{side}$$ Since it's a square, length must equal width. So we first check if $\frac{x-1}{x+1} = \frac{2x}{15}$. 3. **Set length equal to width and solve for $x$:** $$\frac{x-1}{x+1} = \frac{2x}{15}$$ Cross multiply: $$15(x-1) = 2x(x+1)$$ Expand both sides: $$15x - 15 = 2x^2 + 2x$$ Bring all terms to one side: $$0 = 2x^2 + 2x - 15x + 15$$ $$0 = 2x^2 - 13x + 15$$ 4. **Solve quadratic equation $2x^2 - 13x + 15 = 0$ using the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=-13$, $c=15$. Calculate discriminant: $$\Delta = (-13)^2 - 4 \times 2 \times 15 = 169 - 120 = 49$$ Calculate roots: $$x = \frac{13 \pm 7}{4}$$ So, $$x_1 = \frac{13 + 7}{4} = \frac{20}{4} = 5$$ $$x_2 = \frac{13 - 7}{4} = \frac{6}{4} = \frac{3}{2}$$ 5. **Calculate side length for each $x$ value:** For $x=5$: $$\text{side} = \frac{5-1}{5+1} = \frac{4}{6} = \frac{2}{3}$$ For $x=\frac{3}{2}$: $$\text{side} = \frac{\frac{3}{2} - 1}{\frac{3}{2} + 1} = \frac{\frac{1}{2}}{\frac{5}{2}} = \frac{1}{2} \times \frac{2}{5} = \frac{1}{5}$$ 6. **Calculate area for each side length:** For side $= \frac{2}{3}$: $$\text{Area} = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$$ For side $= \frac{1}{5}$: $$\text{Area} = \left(\frac{1}{5}\right)^2 = \frac{1}{25}$$ **Final answer:** The area of the square can be either $\frac{4}{9}$ or $\frac{1}{25}$ depending on the value of $x$.