Question

if x=9-4√5 find √x+1/√x

Please answer fast:)​

Answer 1

Step-by-step explanation:

Here is your answer...✌

x = 9 - 4 \sqrt{5}x=9−4

5

\begin{gathered}\sqrt{x} = \sqrt{9 - 4 \sqrt{5} } \\ \\ \sqrt{x} = \sqrt{ 5 + 4 - 4 \sqrt{5} } \\ \\ \sqrt{x} = \sqrt{ {(\sqrt{5})}^{2} + {2}^{2} - 2 \times \sqrt{5} \times 2 } \\ \\ \sqrt{x} = \sqrt{{ ( \sqrt{5} - 2) }^{2} } \\ \\ \sqrt{x} = \sqrt{5} - 2\end{gathered}

x

=

9−4

5

x

=

5+4−4

5

x

=

(

5

)

2

+2

2

−2×

5

×2

x

=

(

5

−2)

2

x

=

5

−2

And

\begin{gathered}\frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{5} - 2 } \\ \\ \frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{5} - 2} \times \frac{ \sqrt{5} + 2 }{ \sqrt{5} + 2} \\ \\ \frac{1}{ \sqrt{x} } = \frac{ \sqrt{5} + 2 }{{( \sqrt{5})}^{2} - {2}^{2} } \\ \\ \frac{1}{ \sqrt{x} } = \frac{ \sqrt{5} + 2 }{5 - 4} = \frac{ \sqrt{5} + 2}{1} \\ \\ \frac{1}{ \sqrt{x} } = \sqrt{5} + 2\end{gathered}

x

1

=

5

−2

1

x

1

=

5

−2

1

×

5

+2

5

+2

x

1

=

(

5

)

2

−2

2

5

+2

x

1

=

5−4

5

+2

=

1

5

+2

x

1

=

5

+2

Now,

\begin{gathered}\sqrt{x} - \frac{1}{\sqrt{x} } \\ \\ = \sqrt{5} - 2 - \sqrt{5} - 2 \\ \\ = -4\end{gathered}

x

−

x

1

=

5

−2−

5

−2

=−4