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Use the binomial expansion, in ascending powers of $x$, to show that $$ ext{ } \sqrt{(4 - x)} = 2 - \frac{1}{4}x + kx^2 + ...$$ where $k$ is a rational constant to be found - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 2

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Use-the-binomial-expansion,-in-ascending-powers-of-$x$,-to-show-that--$$-ext{-}-\sqrt{(4---x)}-=-2---\frac{1}{4}x-+-kx^2-+-...$$--where-$k$-is-a-rational-constant-to-be-found-Edexcel-A-Level Maths Pure-Question 9-2017-Paper 2.png

Use the binomial expansion, in ascending powers of $x$, to show that $$ ext{ } \sqrt{(4 - x)} = 2 - \frac{1}{4}x + kx^2 + ...$$ where $k$ is a rational constant to... show full transcript

Worked Solution & Example Answer:Use the binomial expansion, in ascending powers of $x$, to show that $$ ext{ } \sqrt{(4 - x)} = 2 - \frac{1}{4}x + kx^2 + ...$$ where $k$ is a rational constant to be found - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 2

Step 1

Use the binomial expansion to show that $\sqrt{(4 - x)} = 2 - \frac{1}{4}x + kx^2 + ...$

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Answer

To begin the expansion, factor out 4 from the square root:
4(1x4)=2(1x4)\sqrt{4(1 - \frac{x}{4})} = 2\sqrt{(1 - \frac{x}{4})}
Using the binomial expansion for (1+u)n(1 + u)^{n}, where u=x4u = -\frac{x}{4} and n=12n = \frac{1}{2}:

\sqrt{(1 - u)} & = 1 - \frac{1}{2}u + \frac{1}{2}\left(\frac{-1}{2}\right)\frac{u^2}{2!} + ... \ & = 1 - \frac{1}{2}\left(-\frac{x}{4}\right) + \frac{1}{2}\left(\frac{-1}{2}\right)\frac{(-\frac{x}{4})^2}{2} + ... \ & = 1 + \frac{x}{8} + \frac{1}{64}x^2 + ... \end{align*}$$ Thus, we have: $$\sqrt{(4 - x)} = 2\left(1 + \frac{x}{8} + \frac{1}{64}x^2 + ...\right)$$ Simplifying gives: $$\sqrt{(4 - x)} = 2 + \frac{1}{4}x - \frac{1}{64}x^2 + ...$$ where $k = -\frac{1}{64}$.

Step 2

State, giving a reason, if the expansion is valid for this value of x.

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Answer

The expansion is valid for x<4|x| < 4, so when substituting x=1x = 1, it is valid because 1<4|1| < 4. Thus, the approximation for 3\sqrt{3} can be obtained using this expansion.

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