If-x-2-x-17-Find-x-2-4-x-2-1-x-2-

Question Number 190034 by Shrinava last updated on 26/Mar/23

If x + \frac{2}{x} = \sqrt{17}

Find : \frac{(x^{2} - 4) (x^{2} - 1)}{x^{2}} = ?

Answered by BaliramKumar last updated on 26/Mar/23

x^{2} + \frac{4}{x^{2}} + 2 \cdot x \cdot \frac{2}{x} = 17

x^{2} + \frac{4}{x^{2}} + 4 = 17

x^{4} + 4 = 13 x^{2} \dots \dots \dots (i)

\frac{(x^{2} - 4) (x^{2} - 1)}{x^{2}} = \frac{x^{4} - 5 x^{2} + 4}{x^{2}} = \frac{(x^{4} + 4) - 5 x^{2}}{x^{2}}

\frac{13 x^{2} - 5 x^{2}}{x^{2}} = \frac{8 x^{2}}{x^{2}} = 8

Answered by Rasheed.Sindhi last updated on 26/Mar/23

If x + \frac{2}{x} = \sqrt{17}

Find : \frac{(x^{2} - 4) (x^{2} - 1)}{x^{2}} = ?

x^{2} = \sqrt{17} x - 2

\frac{(x^{2} - 4) (x^{2} - 1)}{x^{2}}

= \frac{(\sqrt{17} x - 2 - 4) (\sqrt{17} x - 2 - 1)}{\sqrt{17} x - 2}

= \frac{(\sqrt{17} x - 6) (\sqrt{17} x - 3)}{\sqrt{17} x - 2}

= \frac{{17 x}^{2} - 9 \sqrt{17} x + 18}{\sqrt{17} x - 2}

= \frac{17 (\sqrt{17} x - 2) - 9 \sqrt{17} x + 18}{\sqrt{17} x - 2}

= \frac{17 \sqrt{17} x - 34 - 9 \sqrt{17} x + 18}{\sqrt{17} x - 2}

= \frac{8 \sqrt{17} x - 16}{\sqrt{17} x - 2} = \frac{8 (\sqrt{17} x - 2)}{\sqrt{17} x - 2} = 8

Commented by Shrinava last updated on 01/Apr/23

cool professor thankyou

Answered by Rasheed.Sindhi last updated on 26/Mar/23

x + \frac{2}{x} = \sqrt{17}; \frac{(x^{2} - 4) (x^{2} - 1)}{x^{2}} = ?

\frac{(x^{2} - 4) (x^{2} - 1)}{x^{2}} = \frac{x^{2} - 4}{x} \cdot \frac{x^{2} - 1}{x}

= (x - \frac{4}{x}) (x - \frac{1}{x})

= x^{2} + \frac{4}{x^{2}} - 5 = {(x + \frac{2}{x})}^{2} - 4 - 5

= {(\sqrt{17})}^{2} - 9 = 17 - 9 = 8