(a/b)+(b/a)=5 ; (a^2 /b)+(b^2 /a) =12 (1/a)+(1/b)=?


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Question Number 187076 by Humble last updated on 13/Feb/23

$\frac{a}{b} + \frac{b}{a} = 5; \frac{a^{2}}{b} + \frac{b^{2}}{a} = 12$ $\frac{1}{a} + \frac{1}{b} = ?$
	Answered by Rasheed.Sindhi last updated on 13/Feb/23

	$\frac{a}{b} + \frac{b}{a} = 5; \frac{a^{2}}{b} + \frac{b^{2}}{a} = 12; \frac{1}{a} + \frac{1}{b} = ?$ $a^{2} + b^{2} = 5 a b; a^{3} + b^{3} = 12 a b$ ${(a + b)}^{2} = 7 a b; {(a + b)}^{3} = 12 a b + 3 a b (a + b)$ $a + b = \frac{a b (12 + 3 (a + b))}{7 a b}$ $a + b = \frac{12}{7} + \frac{3}{7} (a + b)$ $7 (a + b) - 3 (a + b) = 12$ $a + b = 12 / 4 = 3$ $a b = \frac{{(a + b)}^{2}}{7} = \frac{3^{2}}{7} = \frac{9}{7}$ $\frac{a + b}{a b} = \frac{3}{9 / 7} = 3 \times \frac{7}{9} = \frac{7}{3}$ $\frac{1}{a} + \frac{1}{b} = \frac{7}{3}$
	Commented by BaliramKumar last updated on 13/Feb/23

	$s t e p 3$ $d e v i d w i l l b e 7 a b$
	Commented by Rasheed.Sindhi last updated on 13/Feb/23

	$T h a n X B a l i r a m . I^{'} v e c o r r e c t e d .$
	Commented by Humble last updated on 13/Feb/23

	$t h a n k y o u s i r$
	Answered by HeferH last updated on 13/Feb/23

	$\frac{b}{a} + \frac{b}{a} = 5 \Rightarrow$ $a^{2} + b^{2} = 5 a b$ $\frac{a^{2}}{b} + \frac{b^{2}}{a} = 12 \Rightarrow$ $a^{3} + b^{3} = 12 a b$ $(a + b) (a^{2} + b^{2} - a b) = 12 a b$ $(a + b) (5 a b - a b) = 12 a b$ $a + b = 3$ ${(a + b)}^{2} = 3^{2}$ $a^{2} + b^{2} + 2 a b = 9$ $5 a b + 2 a b = 9$ $a b = \frac{9}{7} \Rightarrow$ $\frac{1}{a} + \frac{1}{b} = \frac{a + b}{a b} = \frac{3}{\frac{9}{7}} = 3 \cdot \frac{7}{9} = \frac{7}{3}$
	Commented by Humble last updated on 13/Feb/23

	$t h a n k y o u, s i r$
	Answered by Rasheed.Sindhi last updated on 13/Feb/23

	$\underset{(i)}{\underset{⏟}{\frac{a}{b} + \frac{b}{a} = 5}}; \underset{(i i)}{\underset{⏟}{\frac{a^{2}}{b} + \frac{b^{2}}{a} = 12}}; \frac{1}{a} + \frac{1}{b} = ?$ $(i) \Rightarrow a^{2} + b^{2} = 5 a b$ $(i i) \Rightarrow a^{3} + b^{3} = 12 a b$ $(i) \Rightarrow a^{2} - a b + b^{2} = 4 a b$ $(a + b) (a^{2} - a b + b^{2} = 4 a b (a + b)$ $a^{3} + b^{3} = 4 a b (a + b)$ $12 a b = 4 a b (a + b)$ $a + b = 3 . . . . (i i i)$ $a^{2} + b^{2} = 5 a b$ $a^{2} + 2 a b + b^{2} = 7 a b$ $a b = \frac{{(a + b)}^{2}}{7} = \frac{3^{2}}{7} = \frac{9}{7} . . . . (i v)$ $(i i i) / (i v) : \frac{a + b}{a b} = 3 / (\frac{9}{7}) = 3 \times \frac{7}{9} = \frac{7}{3}$ $\frac{1}{a} + \frac{1}{b} = \frac{7}{3}$
	Commented by Humble last updated on 14/Feb/23

	$t h a n k y o u, s i r$
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