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2. (a) Express $\sqrt{108}$ in the form $\alpha \sqrt{3}$, where $a$ is an integer - Edexcel - A-Level Maths Pure - Question 4 - 2007 - Paper 2

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2.-(a)-Express-$\sqrt{108}$-in-the-form-$\alpha-\sqrt{3}$,-where-$a$-is-an-integer-Edexcel-A-Level Maths Pure-Question 4-2007-Paper 2.png

2. (a) Express $\sqrt{108}$ in the form $\alpha \sqrt{3}$, where $a$ is an integer. (b) Express $(2 - \sqrt{3})^{2}$ in the form $b + c \sqrt{3}$, where $b$ and $c... show full transcript

Worked Solution & Example Answer:2. (a) Express $\sqrt{108}$ in the form $\alpha \sqrt{3}$, where $a$ is an integer - Edexcel - A-Level Maths Pure - Question 4 - 2007 - Paper 2

Step 1

Express $\sqrt{108}$ in the form $\alpha \sqrt{3}$, where $a$ is an integer.

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Answer

To express 108\sqrt{108} in the required form, we first simplify 108\sqrt{108}. The number can be factored as follows:

108=36⋅3=36⋅3=63.\sqrt{108} = \sqrt{36 \cdot 3} = \sqrt{36} \cdot \sqrt{3} = 6\sqrt{3}.

Thus, we have:

α=6.\alpha = 6.

Therefore, the final answer is 636 \sqrt{3}.

Step 2

Express $(2 - \sqrt{3})^{2}$ in the form $b + c \sqrt{3}$.

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Answer

To expand (2−3)2(2 - \sqrt{3})^{2}, we apply the formula for the square of a binomial:

(a−b)2=a2−2ab+b2.(a - b)^{2} = a^{2} - 2ab + b^{2}.

In our case, a=2a = 2 and b=3b = \sqrt{3}, so:

(2−3)2=22−2(2)(3)+(3)2 (2 - \sqrt{3})^{2} = 2^{2} - 2(2)(\sqrt{3}) + (\sqrt{3})^{2} =4−43+3. = 4 - 4\sqrt{3} + 3.

Now combining the terms gives:

=(4+3)−43=7−43.= (4 + 3) - 4\sqrt{3} = 7 - 4\sqrt{3}.

From this form, it is clear that:

b=7,c=−4.b = 7,\quad c = -4.

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