# Use the binomial expansion to find the square root of 4.8.

## Question:

Use the binomial expansion to find the square root of 4.8.

## Binomial Expansion

The binomial expansion is a mathematical lemma that allows to compute the powers of a binomial expression {eq}(x + y)^n,~~~~ n \in \mathbb{N} {/eq}. Although the original version only applied to integers as an exponent, Newton's generalized binomial theorem expanded the standard binomial theorem to non-integer exponents

## Answer and Explanation:

Newton's generalized binomial theorem expands the standard binomial theorem to non-integer exponents. According to this theorem, the square root of a number {eq}(1 - 2 \, x) {/eq} can be expressed by $$\sqrt{1- 2 \, x} = \sum_{n=0}^\infty {\frac{1}{2} \choose n} \, (- \ 2 \, x)^n = 1 - x - \frac{1}{2} \, x^2 - \frac{1}{2} \, x^3 - \, \cdots~. $$ The first expression under the sum is the binomial coefficient.

Since 4.8 is much larger than 1, we have to take a detour here. We are looking for a number {eq}y {/eq} fulfilling the following condition: $$\begin{align*} \sqrt{1 - 2 \, x} &= \sqrt{4.8 \, y^2} &\text{which would yield} \\[6pt] \sqrt{4.8} &= \frac{1}{y} \, \sqrt{1 - 2 \, x} &(1) \\[6pt] &= \frac{1}{y} \, \left( 1 - x - \frac{1}{2} \, x^2 - \frac{1}{2} \, x^3 - \, \cdots \right) &(2) \end{align*} $$ To ensure validity, we have to ensure that {eq}| x | < \frac{1}{2} {/eq} (convergence of the binomial expression), the smaller {eq}x {/eq}, the better. Also, since (following from (1)) $$\begin{align*} x &= \frac{1 - 4.8 \, y^2}{2}~, &(3) \\[6pt] \end{align*} $$ {eq}y {/eq} should be the quotient of two square numbers, which is close to {eq}\frac{1}{4.8} {/eq} to ensure a small {eq}x {/eq} according to (3). With some try-and-error, we find $$y^2 = \frac{100}{484} = \left( \frac{10}{22} \right)^2 $$ Inserting this into (3), we get $$x = \frac{1 - \frac{4.8}{4.84}}{2} \approx \frac{1 - \frac{1}{1.008}}{2} \approx 0.004 $$ Inserting the obtained values into (2), we get the sought-after root $$\sqrt{4.8} = \frac{22}{10} \, \left( 1 - 0.004 - \frac{1}{2} \, 0.004^2 - \frac{1}{2} \, 0.004^3 - \, \cdots \right) \approx 2.196 $$.

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from Algebra II Textbook

Chapter 21 / Lesson 16