Question

The sum of the reciprocals of two consecutive positive even integers is 11/60 . What are the two numbers

Answer 1

$Let \: x \: and \:(x+2) \: are \: consecutive \\even \: numbers$

$\blue { Reciprocals \: of \: above \: numbers}\\\blue { are \: \frac{1}{x} \:and \: \frac{1}{x+2}}$

/*According to the problem given */

$\pink { \frac{1}{x} + \frac{1}{x+2} = \frac{11}{60}}$

$\implies \frac{x+2+x}{x(x+2)} = \frac{11}{60}$

$\implies \frac{2x+2}{x^{2}+2x} = \frac{11}{60}$

$\implies 60(2x+2) = 11(x^{2}+2x)$

$\implies 120x + 120 = 11x^{2} + 22x$

$\implies 0 = 11x^{2} + 22x - 120x - 120$

$\implies 11x^{2} - 98x - 120= 0$

/* Compare this with ax²+bx+c=0 , we get */

$a = 11, b = -98 , c = -120$

$Discreminant (D) = b^{2} - 4ac$

$= (-98)^{2} - 4 \times 11 \times (-120) \\= 9604+5280\\= 14884$

$\underline { \blue { By \: Quadratic \:Formula:}}$

$x = \frac{-b\pm \sqrt{D}}{2a}$

$\implies x = \frac{-(-98) \pm \sqrt{14884}}{2\times 11} \\= \frac{98\pm 122}{22}$

$\therefore x = \frac{98+ 122}{22}\:Or \: x = \frac{98- 122}{22}$

$\implies x = \frac{220}{22}\:Or \: x = \frac{-24}{22}$

$\implies x = 10\:Or \: x = \frac{-12}{11}$

/* But x is positive even integer */

$\green {x = 10 }$

Therefore.,

$\green { 10\:and \:12 \: are \: required \:two }\\\green { consecutive \:even \: integers }$

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