Question

$a = 9 - 4 \sqrt{5} \: then \: find \: value \: of \: \sqrt{a} \: + 1 \div \sqrt{a}$

Answer 1

$<b>Heya mate. (^_-). Solution below.$
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$<u>Your Question -</u>$

If a = 9 - 4√5, find the value of √a + (1/√a)

$<u>Solution - </u>$

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

$= > \sqrt{a} = \sqrt{9 - 4 \sqrt{5} }$

Breaking 9 into 2 parts

$= > \sqrt{a} = \sqrt{4 + 5 - 4 \sqrt{5} }$

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

Now, using the formula
x² + y² - 2xy = (x+y)²

$= > \sqrt{a} = \sqrt{(2 - \sqrt{5})^{2} }$

$= > \sqrt{a} = (2 - \sqrt{5}) ^{2 \times \frac{1}{2} }$

$= > \sqrt{a} = 2 - \sqrt{5}$

So, √a = 2 - √5

Now,

$\frac{1}{ \sqrt{a} } = \frac{1}{2 - \sqrt{5} }$

Rationalising the denominator,

$= > \frac{1}{ \sqrt{a}} = \frac{1}{2 - \sqrt{5} } \times \frac{2 + \sqrt{5} }{2 + \sqrt{5} }$

$= > \frac{1}{ \sqrt{a} } = \frac{1(2 + \sqrt{5}) }{( {2}^{2}) - ( \sqrt{5} ^{2} )}$

$= > \frac{1}{ \sqrt{a} } = \frac{2 + \sqrt{5} }{4 - 5}$

$= > \frac{1}{ \sqrt{a} } = \frac{2 + \sqrt{5} }{ - 1}$

$= > \frac{1}{ \sqrt{a} } = \frac{ - ( - 2 - \sqrt{5}) }{ - (1)}$

Cancelling the minus sign,

$= > \frac{1}{ \sqrt{a} } = - 2 - \sqrt{5}$

So, (1/√a) = -2 - √5

Now,

√a + (1/√a)

= (2 - √5) + (-2 -√5)

= 2 - √5 - 2 - √5

= -2√5

$<marquee>Hence, when a = 9 - 4√5, then √a + (1/√a) = -2√5</marquee>$
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Thank you. ;-)