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6. (a) Show that $(4 + 3\sqrt{x})^2$ can be written as $16 + k\sqrt{x} + 9x$, where $k$ is a constant to be found - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 2

Question 8

$6.-(a)-Show-that-$(4-+-3\sqrt{x})^2$-can-be-written-as-$16-+-k\sqrt{x}-+-9x$,-where-$k$-is-a-constant-to-be-found-Edexcel-A-Level Maths Pure-Question 8-2007-Paper 2.png$

6. (a) Show that $(4 + 3\sqrt{x})^2$ can be written as $16 + k\sqrt{x} + 9x$, where $k$ is a constant to be found. (b) Find $\int (4 + 3\sqrt{x})^2 dx$.

Worked Solution & Example Answer:6. (a) Show that $(4 + 3\sqrt{x})^2$ can be written as $16 + k\sqrt{x} + 9x$, where $k$ is a constant to be found - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 2

Step 1

Show that $(4 + 3\sqrt{x})^2$ can be written as $16 + k\sqrt{x} + 9x$

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Answer

To prove the equation, we start by expanding the left-hand side:

$(4 + 3\sqrt{x})^2 = 4^2 + 2(4)(3\sqrt{x}) + (3\sqrt{x})^2$

Calculating each term gives:

$= 16 + 24\sqrt{x} + 9x$

From this point, we can see that we can express it in the desired form:

$16 + 24\sqrt{x} + 9x \Rightarrow 16 + k\sqrt{x} + 9x$

Here, we find that $k = 24$ .

Step 2

Find $\int (4 + 3\sqrt{x})^2 dx$

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Answer

Using the result from part (a), we know:

$(4 + 3\sqrt{x})^2 = 16 + 24\sqrt{x} + 9x$

Thus, we need to integrate each term separately:

[ \int (4 + 3\sqrt{x})^2 dx = \int (16 + 24\sqrt{x} + 9x) dx ]

Calculating the integrals:

$\int 16 \, dx = 16x$
$\int 24\sqrt{x} \, dx = 24 \cdot \frac{2}{3}x^{3/2} = 16x^{3/2}$
$\int 9x \, dx = \frac{9}{2}x^2$

Combining these results gives:

[ \int (4 + 3\sqrt{x})^2 dx = 16x + 16x^{3/2} + \frac{9}{2}x^2 + C ]

Where $C$ is the constant of integration.

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