Question

if x=9+4√5,find the value of√x-1/√x​

Answer 1

Answer:

4

Step-by-step explanation:

Given that ;

$\sf x = 9 + 4 \sqrt{5}$

It can be written as ;

$\implies \sf x = 5 + 4 + 4 \sqrt{5} \\ \\ \implies \sf x = ( \sqrt{5} ) {}^{2} + (2) {}^{2} + 2 \times 2 \times \sqrt{5} \\ \\ \implies \sf x = ( \sqrt{5} + 2) {}^{2} \\ \\ \implies \sf \sqrt{x } = \sqrt{5} + 2$

Now, find the value of $\sf \dfrac{1}{\sqrt{x}}$,

$\sf \frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{5} + 2 } \\ \\ \implies \sf \frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{5} + 2} \times \frac{ \sqrt{5} - 2 }{ \sqrt{5} - 2 } \\ \\ \implies \sf \frac{1}{ \sqrt{x} } = \frac{ \sqrt{5} - 2}{( \sqrt{5}) {}^{2} - (2) {}^{2}} \\ \\ \implies \sf \frac{1}{ \sqrt{x} } = \frac{ \sqrt{5} - 2}{5 - 4} \\ \\ \implies \sf \frac{1}{ \sqrt{x} } = \sqrt{5} - 2$

Now,

$\implies \sf \sqrt{x} - \frac{1}{ \sqrt{x} } = \sqrt{5} + 2 - \sqrt{5} + 2 \\ \\ \implies \sf \sqrt{x} - \frac{1}{ \sqrt{x} } =2+2 \\ \\ \boxed{ \bf \sqrt{x} - \frac{1}{ \sqrt{x} } = 4 }$

Answer 2

x = 9 + 4√5

__________ [GIVEN]

• We have to find the value of $\sqrt{x} \: - \: \dfrac{1}{ \sqrt{x} }$

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x = 9 + 4√8

Now.. write 9 in terms of 5 and 4. We can write it like..

=> x = 4 + 5 + 4√5

But we want √5. So, we can also write it like

=> x = (2)² + (√5)² + 4√5

=> x = (2)² + (√5)² + 2 × 2√5

(a + b)² = a² + b² + 2ab

=> x = (2 + √5)²

=> √x = 2 + √5 _______ (eq 1)

______________________________

$\dfrac{1}{\sqrt{x} }$ = $\dfrac{1}{2\:+\:\sqrt{5} }$

=> $\dfrac{1}{2\:+\:\sqrt{5} }$ × $\dfrac{2 \: - \: \sqrt{5} }{2\: - \:\sqrt{5} }$

(a - b) (a + b) = a² - b²

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

=> $\dfrac{2\: - \: \sqrt{5} }{ 4\:-\:5 }$

=> √5 - 2 _______ (eq 2)

_______________________________

$\sqrt{x} \: - \: \dfrac{1}{ \sqrt{x} }$

=> 2 + √5 - (√5 - 2) [From (eq 1) and (eq 2)]

=> 2 + √5 - √5 + 2

=> 4

________________________________

$\sqrt{x} \: - \: \dfrac{1}{ \sqrt{x} }$ is 4

____________ [ANSWER]