1. **Problem:** Expand $\sqrt{9 - 4x}$ in ascending powers of $x$ up to $x^3$ and state the range of $x$ for validity. 2. **Formula:** Use binomial expansion for fractional powers: $$(1 + u)^n = 1 + n u + \frac{n(n-1)}{2!} u^2 + \frac{n(n-1)(n-2)}{3!} u^3 + \cdots$$ valid for $|u| < 1$. 3. **Step 1:** Rewrite $\sqrt{9 - 4x} = 3 \sqrt{1 - \frac{4x}{9}} = 3 (1 - \frac{4x}{9})^{1/2}$. 4. **Step 2:** Let $u = -\frac{4x}{9}$ and $n = \frac{1}{2}$. 5. **Step 3:** Apply binomial expansion: $$ (1 + u)^{1/2} = 1 + \frac{1}{2} u + \frac{\frac{1}{2}(\frac{1}{2} - 1)}{2!} u^2 + \frac{\frac{1}{2}(\frac{1}{2} - 1)(\frac{1}{2} - 2)}{3!} u^3 $$ 6. **Step 4:** Calculate coefficients: - $\frac{1}{2} u = \frac{1}{2} \times -\frac{4x}{9} = -\frac{2x}{9}$ - $\frac{1}{2}(\frac{1}{2} - 1) = \frac{1}{2} \times -\frac{1}{2} = -\frac{1}{4}$ - So second order term coefficient: $\frac{-\frac{1}{4}}{2} = -\frac{1}{8}$ - $u^2 = \left(-\frac{4x}{9}\right)^2 = \frac{16x^2}{81}$ - Second order term: $-\frac{1}{8} \times \frac{16x^2}{81} = -\frac{2x^2}{81}$ 7. **Step 5:** Third order term coefficient: $$ \frac{1}{2} \times \left(-\frac{1}{2}\right) \times \left(-\frac{3}{2}\right) = \frac{3}{8} $$ Divide by $3! = 6$ gives $\frac{3}{8 \times 6} = \frac{1}{16}$ 8. **Step 6:** Third order term: $$ \frac{1}{16} \times \left(-\frac{4x}{9}\right)^3 = \frac{1}{16} \times -\frac{64 x^3}{729} = -\frac{4 x^3}{729} $$ 9. **Step 7:** Combine terms: $$ (1 + u)^{1/2} = 1 - \frac{2x}{9} - \frac{2x^2}{81} - \frac{4x^3}{729} $$ Multiply by 3: $$ \sqrt{9 - 4x} = 3 - \frac{2x}{3} - \frac{2x^2}{27} - \frac{4x^3}{243} $$ 10. **Step 8:** Range of validity: Since $|u| = \left| -\frac{4x}{9} \right| < 1$, we have $$ |x| < \frac{9}{4} = 2.25 $$ **Final answer:** $$ \sqrt{9 - 4x} \approx 3 - \frac{2x}{3} - \frac{2x^2}{27} - \frac{4x^3}{243}, \quad |x| < 2.25 $$