Question

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

Answer 1

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.

Answer 2

Given that , x - 1 / x = 5 .

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

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

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

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

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

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

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

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

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

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

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

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

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

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