Question

Q8,9 heeeeeeeeeeeeeeeeeeeeelp

Answer 1

Answer :


Question no. 8

$8. \: ( {x}^{2} + \frac{1}{ {x}^{2} } ) - 4(x + \frac{1}{x} ) + 6 \\ \\ = > \: {x}^{2} + \frac{1}{ {x}^{2} } - 4x - \frac{4}{x } + 6 \\ \\ = > {x}^{2} + \frac{1}{x {}^{2} } - 4x - \frac{4x}{x {}^{2} } + 6 \\ \\ = > \frac{ {x}^{2} \times {x}^{2} }{ {x}^{2} } + \frac{1}{ {x}^{2} } - \frac{x {}^{2} \times 4x}{ {x}^{2} } - \frac{4x}{ {x}^{2} } + \frac{ {x}^{2} \times 6}{ {x}^{2} } \\ \\ = > \frac{ {x}^{4} }{ {x}^{2} } + \frac{1}{ {x}^{2} } - \frac{ {4x}^{3} }{ {x}^{2} } - \frac{4x}{x {}^{2} } + \frac{6x {}^{2} }{x {}^{2} } \\ \\ = > \frac{x {}^{4} + 1 - 4x {}^{3} - 4x + 6x {}^{2} }{ {x}^{2} } \\ \\ = > \frac{ {x}^{4} - {4x}^{3} + {6x}^{2} - 4x + 1 }{ {x}^{2} }$




Question no. 9th ---

$9. \: \: (x + \frac{1}{x} ) {}^{2} + 6(x + \frac{1}{x} ) + 9 \\ \\ = > identity \: = > \\ \\ = > \: {a}^{2} + 2ab + {b}^{2} = (a + b) {}^{2} \\ \\ = > factorise \\ \\ = > (x + \frac{1}{x} ) {}^{2} + 2(x + \frac{1}{x} )(3) + {3}^{2} \\ \\ = > (x + \frac{1}{x} + 3) {}^{2}$



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