Math, asked by sumansindhu115, 1 month ago

7. If x-1/x = 5, then the value of x?+1/x? equals to
(1 Point)
13
17
25
27​

Answers

Answered by amansharma264
68

EXPLANATION.

⇒ (x - 1/x) = 5.

As we know that,

Squaring on both sides of the equation, we get.

⇒ (x - 1/x)² = (5)².

⇒ x² + 1/x² - 2(x)(1/x) = 25.

⇒ x² + 1/x² - 2 = 25.

⇒ x² + 1/x² = 25 + 2.

⇒ x² + 1/x² = 27.

Hence, option [D] is correct answer.

Answered by BrainlyRish
62

Given that , x - 1 / x = 5 .

Exigency To Find : Value of x² + 1 /x² ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀¤ Finding Value of + 1 /x² :

\qquad \dashrightarrow \sf x - \dfrac{  1\:}{x} \:\:=\:5 \:\\\\\qquad \bigstar  \:\:\sf On  \:Squaring \:both \:sides \: we \:get \:, \:\\\\\qquad \dashrightarrow \sf x - \dfrac{ 1\:}{x} \:\:=\:5 \:\\\\\qquad \dashrightarrow \sf \Big\{x - \dfrac{  1\:}{x} \Big\}^2 \:\:=\:(5)^2 \:\\\\\qquad \dashrightarrow \sf \Big\{x - \dfrac{  1\:}{x} \Big\}^2 \:\:=\:25 \:\\\\\qquad \dashrightarrow \sf x^2 + \dfrac{ 1\:}{x^2 } - 2 \times x \times \dfrac{1}{x}  \:\:=\:25 \:\\\\\qquad \dashrightarrow \sf x^2 + \dfrac{ 1\:}{x^2 } - 2 \times \cancel{x} \times \dfrac{1}{\cancel{x}}  \:\:=\:25 \:\\\\\qquad \dashrightarrow \sf x^2 + \dfrac{  1\:}{x^2 } - 2  \:\:=\:25 \:\\\\\qquad \dashrightarrow \sf x^2 + \dfrac{  1\:}{x^2 }  \:\:=\:25 + 2  \:\\\\\qquad \dashrightarrow \sf x^2 + \dfrac{  1\:}{x^2 }  \:\:=\:27  \:\\\\\dashrightarrow \underline {\boxed{\pmb{\frak{\purple { x^2 +  \dfrac{  1\:}{x^2 }  \:\:=\:27}}}}}\:\:\bigstar \\\\

\qquad \therefore \underline {\sf Hence,  \:Value \:of \:x^2 +1 / x^2 \:is\:{\bf  27\:}.\:}\\

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

\qquad \qquad \underline {\bigstar\pmb{\mathbb{ ADDITIONAL \:\:INFORMATION \::\:}}}\:\\\\

\dag \:\:\underline { \underline {\purple{\sf Algebraic \; Indentity \:\:-\: }}}\\\\

\qquad \sf ( I ) \:\:( a + b)^2 =\:a^2 + b^2 + 2ab \:\\\\

\qquad \sf ( II ) \:\:( a - b)^2 =\:a^2 + b^2 - 2ab \:\\\\

\qquad \sf ( III ) \:\: a^2 - b^2 =\:( a + b ) ( a - b ) \:\\\\

\qquad \sf ( IV ) \:\:( x + b ) ( x + b ) \:=\:x^2 + ( a + b ) x + ab \:\\\\

\qquad \sf ( V ) \:\: ( a + b + c )^2\:=\: a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \: \:\\\\

\qquad \sf ( VI ) \:\: ( a + b  )^3\:=\: a^3 + b^3  + 3ab ( a + b )  \: \:\\\\

\qquad \sf ( VII ) \:\: ( a - b  )^3\:=\: a^3 - b^3  - 3ab ( a - b )  \: \:\\\\

\qquad \sf ( VII ) \:\: a^3 + b^3 + c^3 - 3abc\:=\: ( a + b + c ) \: ( a^2 + b^2 + c^2 - ab - bc - ca  )   \: \:\\\\

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