If x = 5−265 - 2\sqrt{6}5−26, find the value of x2+1x2x^2 + \dfrac{1}{x^2}x2+x21.
25 Likes
Given x = 5−265 - 2\sqrt{6}5−26
∴1x=15−26=15−26×5+265+26⇒5+26(5)2−(26)2⇒5+2625−24=5+261⇒1x=5+26∴(x+1x)=(5−26)+(5+26)=10….(i)\therefore \dfrac{1}{x} = \dfrac{1}{5 - 2\sqrt{6}} = \dfrac{1}{5 - 2\sqrt{6}} × \dfrac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} \\[1.5em] \Rightarrow \dfrac{5 + 2\sqrt{6}}{(5)^2 - (2\sqrt{6})^2} \\[1.5em] \Rightarrow \dfrac{5 + 2\sqrt{6}}{25 - 24} = \dfrac{5 + 2\sqrt{6}}{1} \\[1.5em] \Rightarrow\dfrac{1}{x} = 5+2\sqrt6 \\[1.5em] \therefore (x + \dfrac{1}{x}) = (5 - 2\sqrt6) + (5 + 2\sqrt6) = 10 \qquad \text{….(i)}∴x1=5−261=5−261×5+265+26⇒(5)2−(26)25+26⇒25−245+26=15+26⇒x1=5+26∴(x+x1)=(5−26)+(5+26)=10….(i)
We know that (x+1x)2=x2+1x2+2{\Big(x + \dfrac{1}{x}\Big)}^2 = x^2 + \dfrac{1}{x^2} + 2(x+x1)2=x2+x21+2
⇒x2+1x2=(x+1x)2−2⇒x2+1x2=102−2…..using(i)⇒x2+1x2=100−2x2+1x2=98\Rightarrow x^2 + \dfrac{1}{x^2} = {\Big(x + \dfrac{1}{x}\Big)}^2 -2 \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = 10^2 - 2 ….. \text{using(i)} \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = 100 - 2 \\[1.5em] \bold{x^2 + \dfrac{1}{x^2} = 98}⇒x2+x21=(x+x1)2−2⇒x2+x21=102−2…..using(i)⇒x2+x21=100−2x2+x21=98
Answered By
15 Likes
If a = 2 + 3\sqrt{3}3 , then find the value a−1aa - \dfrac{1}{a}a−a1.
If x=1−2x = 1-\sqrt{2}x=1−2, find the value of (x−1x)4{\Big(x-\dfrac{1}{x}\Big)}^4(x−x1)4.
If p = 2−52+5\dfrac{2-\sqrt{5}}{2+\sqrt{5}}2+52−5 and q = 2+52−5\dfrac{2+\sqrt{5}}{2-\sqrt{5}}2−52+5 find the values of :
(i) p + q
(ii) p - q
(iii) p2 + q2
(iv) p2 - q2
If x 2−12+1\dfrac{\sqrt{2}-1}{\sqrt{2}+1}2+12−1 and y = 2+12−1\dfrac{\sqrt{2}+1}{\sqrt{2}-1}2−12+1 , then find the value of x2+5xy+y2{x^2 +5xy + y^2}x2+5xy+y2.
