Question

find the value of :-
$x { }^{2} + \frac{1}{x} \: when \: x + \frac{1}{x} = 6$
​

Answer 1

Answer:

$Using \: \: the \: \: identity: \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

Step-by-step explanation:

$x + \frac{1}{x} = 6 \\ {(x + \frac{1}{x} )}^{2} = {6}^{2} \\ {x}^{2} + \frac{1}{ {x}^{2} } = 36 \\ {x}^{2} + \frac{1}{ {x}^{2} } + (2 \times \cancel x \times \frac{1}{ \cancel x} ) = 36 \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 36 \\ {x}^{2} + \frac{1}{ {x}^{2} } = 36 - 2 \\ \boxed{ \large {x}^{2} + \frac{1}{ {x}^{2} } = \underline{ \underline{ 34}}}$