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(a) Write $\sqrt{45}$ in the form $\alpha\sqrt{5}$, where $\alpha$ is an integer - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 1

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(a)-Write-$\sqrt{45}$-in-the-form-$\alpha\sqrt{5}$,-where-$\alpha$-is-an-integer-Edexcel-A-Level Maths Pure-Question 7-2006-Paper 1.png

(a) Write $\sqrt{45}$ in the form $\alpha\sqrt{5}$, where $\alpha$ is an integer. (b) Express $\frac{2(3 + \sqrt{5})}{(3 - \sqrt{5})}$ in the form $b + c\sqrt{5}$, ... show full transcript

Worked Solution & Example Answer:(a) Write $\sqrt{45}$ in the form $\alpha\sqrt{5}$, where $\alpha$ is an integer - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 1

Step 1

Write $\sqrt{45}$ in the form $\alpha\sqrt{5}$, where $\alpha$ is an integer.

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Answer

To simplify 45\sqrt{45}, we start by factorizing 45:

45=9×5=32×5.45 = 9 \times 5 = 3^2 \times 5.

Thus, we can express 45\sqrt{45} as:

45=9×5=95=35.\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \cdot \sqrt{5} = 3\sqrt{5}.

Therefore, in the form α5\alpha\sqrt{5}, we have α=3\alpha = 3.

Step 2

Express $\frac{2(3 + \sqrt{5})}{(3 - \sqrt{5})}$ in the form $b + c\sqrt{5}$, where $b$ and $c$ are integers.

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Answer

To simplify 2(3+5)(35)\frac{2(3 + \sqrt{5})}{(3 - \sqrt{5})}, we first multiply the numerator and denominator by the conjugate of the denominator, which is (3+5)(3 + \sqrt{5}):

2(3+5)(3+5)(35)(3+5).\frac{2(3 + \sqrt{5})(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})}.

Calculating the denominator:

(35)(3+5)=32(5)2=95=4.(3 - \sqrt{5})(3 + \sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4.

Now for the numerator:

2(3+5)(3+5)=2(9+35+35+5)=2(14+65)=28+125.2(3 + \sqrt{5})(3 + \sqrt{5}) = 2(9 + 3\sqrt{5} + 3\sqrt{5} + 5) = 2(14 + 6\sqrt{5}) = 28 + 12\sqrt{5}.

Thus, we can rewrite the expression as:

28+1254=7+35.\frac{28 + 12\sqrt{5}}{4} = 7 + 3\sqrt{5}.

Therefore, in the form b+c5b + c\sqrt{5}, we have b=7b = 7 and c=3c = 3.

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