Question

The sum of a number and its recipocal.is 52 /21 find the number

Answer 1

» let's the required no. be X

» then its reciprocal is = 1/X

Now

» according to the question :-

$= > x + \frac{1}{x} = \frac{52}{21} \\ \\ = > \frac{ {x}^{2} + 1}{x} = \frac{52}{21} \\ \\ = > 21 {x}^{2} + 21 = 52x \\ \\ = > 21 {x}^{2} - 52x + 21 = 0$
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» Using factorise formula :-

$= > x = \frac{ - b \binom{ + }{ - } \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ = > x = \frac{ - ( - 52) \binom{ + }{ - } \sqrt{ { (- 52)}^{2} - 4(21)(21) } }{2 \times 21} \\ \\ = > x = \frac{52\binom{ + }{ - } \sqrt{2704 - 1764} }{42} \\ \\ = > x = \frac{52 \binom{ + }{ - } \sqrt{940} }{42}$

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so

$= > \frac{1}{x} = \frac{42}{52 \binom{ + }{ - } \sqrt{ 940} }$

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Answer 2

• Let the number be M

» The sum of a number and its recipocal is 52/21.

• Reciprocaled number = 1/M

A.T.Q.

=> $M\:+\:\dfrac{1}{M}\:=\:\dfrac{52}{21}$

=> $\dfrac{ {M}^{2} \: + \: 1}{M} \: = \: \dfrac{52}{21}$

Cross-multiply them

=> $21({M}^{2} \: + \: 1)\:=\:52(M)$

=> $21{M}^{2} \: + \: 21\:=\:52M$

=> $21{M}^{2} \:-\:52M\: + \: 21\:=\:0$

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Now.. solve by using quadratic formula

i.e

Here..

x = M

» $M \: = \: \dfrac{ - b \: \pm \: \sqrt{( {b)}^{2} \: - \: 4ac} }{2a}$

Put value of a, b and c in above formula

=> $M \: = \: \dfrac{ - ( - 52) \: \pm \: \sqrt{( { - 52)}^{2} \: - \: 4(21)(21)} }{2(21)}$

=> $M \: = \: \dfrac{ 52 \: \pm \: \sqrt{2704 \: - \: 1764} }{42}$

=> $M \: = \: \dfrac{ 52 \: \pm \: \sqrt{940} }{42}$

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Number = $M\: = \: \dfrac{ 52 \: \pm \: \sqrt{940} }{42}$

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