Question

if
$if \: 5 + 2\sqrt{2} \: find \: a + \frac{1}{a}$
​

Answer 1

Answer:

Hello friends!!

\frac{ \sqrt{5} - 2}{ \sqrt{5} + 2} - \frac{ \sqrt{5} + 2}{ \sqrt{5} - 2 } = a + b \sqrt{5}

5

+2

5

−2

−

5

−2

5

+2

=a+b

5

First we have to rationalise the denominator.

\frac{ \sqrt{5} - 2 }{ \sqrt{5} + 2 } \times \frac{ \sqrt{5} - 2}{ \sqrt{5} - 2 } - \frac{ \sqrt{5} + 2 }{ \sqrt{5} - 2} \times \frac{ \sqrt{5} + 2}{ \sqrt{5} + 2} = a + b \sqrt{5}

5

+2

5

−2

×

5

−2

5

−2

−

5

−2

5

+2

×

5

+2

5

+2

=a+b

5

$$\frac{ (\sqrt{5} - 2)( \sqrt{5} - 2) }{( \sqrt{5} + 2)( \sqrt{5} - 2) } - \frac{( \sqrt{5} + 2)( \sqrt{5} + 2) }{( \sqrt{5} -2)( \sqrt{5} + 2)} = a + b \sqrt{5}$$

$$\frac{ {( \sqrt{5} - 2)}^{2} }{( \sqrt{5} - 2)( \sqrt{5} + 2) } - \frac{ {( \sqrt{5} + 2)}^{2} }{( \sqrt{5} + 2)( \sqrt{5} + 2) } = a + b \sqrt{5}$$

Using identity:

( a - b )² = a² + b² - 2ab

( a + b )² = a² + b² + 2ab

( a - b )( a + b ) = a² - b²

$$\frac{ {( \sqrt{5} )}^{2} + {(2)}^{2} - 2 \times 2 \times \sqrt{5} }{ {( \sqrt{5} )}^{2} - {(2)}^{2} } - \frac{ {( \sqrt{5} )}^{2} + {(2)}^{2} + 2 \times 2 \times \sqrt{5} }{ {( \sqrt{5} )}^{2} - {(2)}^{2} } = a + b \sqrt{5}$$

$$\frac{5 + 4 - 4 \sqrt{5} }{5 - 4} - \frac{5 + 4 + 4 \sqrt{5} }{5 - 4} = a + b \sqrt{5}$$

$$\frac{9 - 4 \sqrt{5} }{1} - \frac{9 + 4 \sqrt{5} }{1} = a + b \sqrt{5}$$

$$9 - 4 \sqrt{5} - (9 + 4\sqrt{5} ) = a + b \sqrt{5}$$

$$9 - 4 \sqrt{5} - 9 - 4 \sqrt{5} = a + b \sqrt{5}$$

$$- 8 \sqrt{5} = a + b \sqrt{5}$$

Comparing these values,

a = 0

b = - 8

HOPE IT HELPS YOU...

Step-by-step explanation:

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