2. (a) Use the binomial theorem to expand $(8-3x)^{\frac{1}{2}}$, $|x| < \frac{3}{8}$, in ascending powers of $x$, up to and including the term in $x^{3}$, giving each term as a simplified fraction. (b) Use your expansion, with a suitable value of $x$, to obtain an approximation to $\\sqrt{(7.7)}$. Give your answer to 7 decimal places.

A-Level Edexcel Maths Pure Integration: 2. (a) Use the binomial theorem

2. (a) Use the binomial theorem to expand $(8-3x)^{\frac{1}{2}}$, $|x| < \frac{3}{8}$, in ascending powers of $x$, up to and including the term in $x^{3}$, giving each term as a simplified fraction - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 8

Question 4

$2.-(a)-Use-the-binomial-theorem-to-expand------$(8-3x)^{\frac{1}{2}}$,-------$|x|-<-\frac{3}{8}$,-----in-ascending-powers-of-$x$,-up-to-and-including-the-term-in-$x^{3}$,-giving-each-term-as-a-simplified-fraction-Edexcel-A-Level Maths Pure-Question 4-2008-Paper 8.png$

2. (a) Use the binomial theorem to expand $(8-3x)^{\frac{1}{2}}$, $|x| < \frac{3}{8}$, in ascending powers of $x$, up to and including the term in $x^... show full transcript

Worked Solution & Example Answer:2. (a) Use the binomial theorem to expand $(8-3x)^{\frac{1}{2}}$, $|x| < \frac{3}{8}$, in ascending powers of $x$, up to and including the term in $x^{3}$, giving each term as a simplified fraction - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 8

Step 1

Use the binomial theorem to expand $(8-3x)^{\frac{1}{2}}$

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Answer

To expand $(8-3x)^{\frac{1}{2}}$ , we first factor out the $8$ :

$(8(1 - \frac{3x}{8}))^{\frac{1}{2}} = 2\sqrt{2}(1 - \frac{3x}{8})^{\frac{1}{2}}$

Next, we apply the binomial expansion formula for $(1 + u)^{n}$ :

$(1 + u)^{n} = 1 + nu + \frac{n(n-1)}{2!}u^{2} + \frac{n(n-1)(n-2)}{3!}u^{3} + \cdots$

In our case, $n = \frac{1}{2}$ and $u = -\frac{3x}{8}$ . Thus, the terms become:

Zero term:
$1$
First term: $-\frac{3}{16}x$
Second term:
$\frac{(\frac{1}{2})(-\frac{3}{8})}{2} x^{2} = -\frac{9}{128}x^{2}$
Third term:
$\frac{(\frac{1}{2})(-\frac{1}{2})(-\frac{3}{8})^{2}}{6} x^{3 = \frac{27}{3072}x^{3}}$

Combining these, the expansion up to $x^{3}$ is:

$2\sqrt{2}\left( 1 - \frac{3}{16}x - \frac{9}{128}x^{2} + \frac{27}{3072}x^{3} \right)$

Thus, the final result is:

$2\sqrt{2} - \frac{3\sqrt{2}}{8}x - \frac{9\sqrt{2}}{64}x^{2} + \frac{27\sqrt{2}}{1536}x^{3}$

Step 2

Use your expansion, with a suitable value of $x$, to obtain an approximation to $\\sqrt{(7.7)}$

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Answer

Select an appropriate value for $x$ that satisfies $|x| < \frac{3}{8}$ . A reasonable choice is:

$x = 0.1.$

Next, substitute $x = 0.1$ into the expansion:

$\\sqrt{(7.7)} \approx 2\sqrt{2} - \frac{3\sqrt{2}}{8}(0.1) - \frac{9\sqrt{2}}{64}(0.1)^{2} + \frac{27\sqrt{2}}{1536}(0.1)^{3}$

Now calculate each term:

First term: $2\sqrt{2} \approx 2 \times 1.4142136 = 2.8284271$
Second term: $-\frac{3\sqrt{2}}{8}(0.1) = -\frac{3(1.4142136)}{8}(0.1) \approx -0.053033$
Third term: $-\frac{9\sqrt{2}}{64}(0.01) = -\frac{9(1.4142136)}{64}(0.01) \approx -0.00200098$
Fourth term: $\frac{27\sqrt{2}}{1536}(0.001) = \frac{27(1.4142136)}{1536}(0.001) \approx 0.0000246085$

Now, summing these values:

$2.8284271 - 0.053033 - 0.00200098 + 0.0000246085 \approx 2.77341773$

Thus, the approximation for $\\sqrt{(7.7)}$ is approximately:

$1.9746809$

Rounded to 7 decimal places, the final result is:

$1.9746809$

2. (a) Use the binomial theorem to expand $(8-3x)^{\frac{1}{2}}$, $|x| < \frac{3}{8}$, in ascending powers of $x$, up to and including the term in $x^{3}$, giving each term as a simplified fraction - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 8

Worked Solution & Example Answer:2. (a) Use the binomial theorem to expand $(8-3x)^{\frac{1}{2}}$, $|x| < \frac{3}{8}$, in ascending powers of $x$, up to and including the term in $x^{3}$, giving each term as a simplified fraction - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 8

Use the binomial theorem to expand $(8-3x)^{\frac{1}{2}}$

Use your expansion, with a suitable value of $x$, to obtain an approximation to $\\sqrt{(7.7)}$

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