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Show that $$\frac{\sqrt{180} - 2\sqrt{5}}{\sqrt{5} - 5}$$ can be written in the form $a + \frac{\sqrt{5}}{b}$ where $a$ and $b$ are integers. - Edexcel - GCSE Maths - Question 21 - 2020 - Paper 1

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Question 21

Show-that-$$\frac{\sqrt{180}---2\sqrt{5}}{\sqrt{5}---5}$$-can-be-written-in-the-form-$a-+-\frac{\sqrt{5}}{b}$-where-$a$-and-$b$-are-integers.-Edexcel-GCSE Maths-Question 21-2020-Paper 1.png

Show that $$\frac{\sqrt{180} - 2\sqrt{5}}{\sqrt{5} - 5}$$ can be written in the form $a + \frac{\sqrt{5}}{b}$ where $a$ and $b$ are integers.

Worked Solution & Example Answer:Show that $$\frac{\sqrt{180} - 2\sqrt{5}}{\sqrt{5} - 5}$$ can be written in the form $a + \frac{\sqrt{5}}{b}$ where $a$ and $b$ are integers. - Edexcel - GCSE Maths - Question 21 - 2020 - Paper 1

Step 1

Process to rationalizing the denominator

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Answer

First, we will simplify the denominator by rationalizing it. To do this, multiply both the numerator and the denominator by the conjugate of the denominator, which is 5+5\sqrt{5} + 5:

(180−25)(5+5)(5−5)(5+5)\frac{(\sqrt{180} - 2\sqrt{5})(\sqrt{5} + 5)}{(\sqrt{5} - 5)(\sqrt{5} + 5)}

The denominator simplifies to:

(5)2−(5)2=5−25=−20(\sqrt{5})^2 - (5)^2 = 5 - 25 = -20

Step 2

Expanding the numerator

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Answer

Next, we will expand the numerator:

180⋅5+5180−25⋅5−105\sqrt{180} \cdot \sqrt{5} + 5\sqrt{180} - 2\sqrt{5} \cdot \sqrt{5} - 10\sqrt{5}

This becomes:

900+5180−10−105\sqrt{900} + 5\sqrt{180} - 10 - 10\sqrt{5}

Or:

30+5180−10−105=20+5180−10530 + 5\sqrt{180} - 10 - 10\sqrt{5} = 20 + 5\sqrt{180} - 10\sqrt{5}

Step 3

Final Simplification

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Answer

Next, we simplify 180\sqrt{180}:

180=36⋅5=65\sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5}

Thus, substituting this back into the equation gives:

20+5⋅65−105−20=20+305−105−20=20+205−20\frac{20 + 5 \cdot 6\sqrt{5} - 10\sqrt{5}}{-20} = \frac{20 + 30\sqrt{5} - 10\sqrt{5}}{-20} = \frac{20 + 20\sqrt{5}}{-20}

We can separate the fractions:

20−20+205−20=−1−5\frac{20}{-20} + \frac{20\sqrt{5}}{-20} = -1 - \sqrt{5}

Thus we can express it as:

−1+5−1-1 + \frac{\sqrt{5}}{-1}

Here, we have a=−1a = -1 and b=−1b = -1, both integers.

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