Question

The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Answer 1

Let Rehman's present age = x yrs

a/c to question,The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3.
e.g., 1/(x - 3) + 1/(x +5) = 1/3

=> {(x + 5) + (x - 3)}/(x - 3)(x + 5) = 1/3

=> (2x + 2)/(x² + 2x - 15) = 1/3

=> 3(2x + 2) = x² + 2x - 15

=> 6x + 6 = x² + 2x - 15

=> x² - 4x - 21 = 0

=> x² - 7x + 3x - 21 = 0

=> x(x - 7) + 3(x - 7) = 0

=> (x - 7)(x + 3) = 0

=> x =- 3 and 7 , but x ≠ -3

hence, Rehman's present age is 7 yrs

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Assume the age of Reham be m

★As per the situation we have :-

${\boxed{\sf\:{\dfrac{1}{m-3}+\dfrac{1}{m+5}=\dfrac{1}{3}}}}$

★After cross multiplication we get :-

3(m + 5) + 3(m - 3) = (m - 3)(m + 5)

3m + 15 + 3m - 9 = m² + 2m - 15

m² - 4m - 21 = 0

(m - 7)(m + 3) = 0

m - 7 = 0

m = 7

m + 3 = 0

m = -3

★As we know that age can't be in negative.

★Hence,

${\boxed{\sf\:{Rehman\;present\;age\;is\;7\;years}}}$

$\boxed{\begin{minipage}{14 cm} Additional Information \\ \\ \ A\; Quadratic\; Equation\;has\;three\;equal\;roots \\ \\ 1)Real\;and\;Distinct \\ \\ 2)Real\;and\;Coincident \\ \\ 3) Imaginary \\ \\ Note:-Third\; Imaginary\;is\;not\;taken\;in\;class\;10th \\ \\ If\;p(x)\;is\;a\; quadratic\; polynomial\;then\;p(x)=0\;is\;called\; Quadratic\; Polynomial \\ \\ General\;Formula=ax^2+bx+c=0 \\ \\ A polynomial\;whose\;degree\;will\;be\;2\;is\; considered\;as\; Quadratic\; Polynomial \\ \\ Rules\;for\; solving\; Quadratic\; Equations:- \\ \\ Put\;all\;the\;terms\;into\;RHS\;and\;make\it\;zero \\ \\ Substitute\;all\; factors\;equal\;to\;Zero\;Get\;a\;equal\; solution \end{minipage}}$