Question

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

Answer 1

x=3+8–√…(1)x=3+8…(1)

x=3+8–√…(1)x=3+8…(1)⇒1x=3−8√(3)2−(8√)2⇒1x=3−8(3)2−(8)2

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)

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=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=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

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=6Now,

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=6Now,x2+1x2x2+1x2

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=6Now,x2+1x2x2+1x2=(x+1x)2–2.x1x=(x+1x)2–2.x1x

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=6Now,x2+1x2x2+1x2=(x+1x)2–2.x1x=(x+1x)2–2.x1x=62–2=62–2