Question

If x = 9- 4root5, find the value of

a) rootx + 1/ rootx
b) x^2 + 1/x^2

Pls. do answer its important ​

Answer 1

Step-by-step explanation:

(i)x = $9-4\sqrt{5}$

$9-4\sqrt{5} =4+5-2*2*\sqrt{5} \\(2)^{2} -2(2)(\sqrt{5} )+(\sqrt{5} )^{2} \\=(2-\sqrt{5} )^{2}$

Therefore x = $(2-\sqrt{5} )^{2}$

So, √x will be $2-\sqrt{5}$

1/√x = $\frac{1}{2-\sqrt{5} } *\frac{2+\sqrt{5} }{2+\sqrt{5} }$

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

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

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

= -2-√5

$\sqrt{x} +\frac{1}{\sqrt{x} } = 2-\sqrt{5} +(-2-\sqrt{5} )$

=  $2-2-\sqrt{5} -\sqrt{5}$

= $-2\sqrt{5}$

(ii) $x^{2} = (9-4\sqrt{5} )^{2}$

= $81-2*9*4\sqrt{5} +80$

= 161 - 72√5

$\frac{1}{x^{2} } =\frac{1}{161-72\sqrt{5} } *\frac{161+72\sqrt{5} }{161+72\sqrt{5} }$

= $\frac{161+72\sqrt{5} }{25921-25920}$

= 161+72√5

Therefore , $x^{2} +\frac{1}{x^{2} } = 161-72\sqrt{5} +161+72\sqrt{5}$

=322

Hence, (i) Answer = -2√5

            (ii) Answer = 322

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