CBSE BOARD XII, asked by virtuoso07, 2 months ago

if x = 3 + \sqrt{8} then the value of ( {x}^{2} + 1 \div {x}^{2} )is​

Answers

Answered by Dik24
6

go through the pic above

hope it helps

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Answered by ayanzubair
11

x=3+8–√…(1)x=3+8…(1)

⇒1x=3−8√(3)2−(8√)2⇒1x=3−8(3)2−(8)2

⇒1x=3−8–√….(2)⇒1x=3−8….(2)

Now adding (1) & (2) we get…

x+1x=(3+8–√)+(3−8–√)x+1x=(3+8)+(3−8)

⇒x+1x=6⇒x+1x=6

Now,

x2+1x2x2+1x2

=(x+1x)2–2.x1x=(x+1x)2–2.x1x

=62–2=62–2

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