Question

If x + 1/x = 3 , then x^2 + 1/x^2 = ?​

Answer 1

Answer:

7

Step-by-step explanation:

if x+1/x =3 , then

(x+ 1/x)² =3²

[according to identity - (a+ 1/a)² =a²+ 1/a² +2 ]

,

x² +1/x² +2 = 9

x² +1/x² = 9-2 = 7

Answer 2

GIVEN

$\\ \red \bigstar \displaystyle \tt \: x + \dfrac{1}{x} = 3$

TO FIND :-

$\\ \red \bigstar \: \displaystyle \tt value \: of \: \: \: x ^{2} + \dfrac{1}{x ^{2} } = 3$

SOLUTION :-

$\\ \longmapsto \displaystyle \tt \: x + \dfrac{1}{x} = 3$

$\orange \bigstar \blue {\displaystyle \: \tt Squaring \: \: both \: \: sides .\: }$

$\longmapsto \: \displaystyle \tt \bigg(\: x + \dfrac{1}{x } \bigg) ^{2} =( 3) ^{2}$

$\orange \bigstar \blue {\displaystyle \: \tt Using \: identity :- (a + b)^{2} = a^{2} + b^{2} + 2ab }$

$\longmapsto \: \displaystyle \tt value \: of \: \: \: (x )^{2} + \bigg( \dfrac{1}{x } \bigg)^{2} + 2 \times \cancel x \times \dfrac{1}{ \cancel x} = 9$

$\longmapsto \: \: \displaystyle \tt value \: of \: \: \: x ^{2} + \dfrac{1}{x ^{2} } + 2 = 9$

$\longmapsto \: \: \displaystyle \tt value \: of \: \: \: x ^{2} + \dfrac{1}{x ^{2} } = 9 - 2$

$\red \bigstar \: \boxed{\displaystyle \tt value \: of \: \: \: x ^{2} + \dfrac{1}{x ^{2} } = 7} \: \red \bigstar$

$\underline{ \underline{ \displaystyle \: \tt \therefore \:Hence \: the \: required \: value \: is \: 7.}}$

Some Identities :-

$\: \blue \bigstar \displaystyle \tt (a + b)^{2} = a^{2} + b^{2} + 2ab$

$\: \blue \bigstar \displaystyle \tt (a - b)^{2} = a^{2} - b^{2} + 2ab$