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Mathematics

The sum of the reciprocals of Joseph's age 3 years ago and five years from now is 13\dfrac{1}{3}. Find his present age.

Quadratic Equations

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Answer

Let present age of Joseph be x years.

It is given that the sum of the reciprocals of Joseph's age 3 years ago and five years from now is 13\dfrac{1}{3}.

⇒1x−3+1x+5=13⇒(x+5)+(x−3)(x−3)(x+5)=13⇒2x+2(x−3)(x+5)=13⇒3(2x+2)=(x−3)(x+5)⇒6x+6=x(x+5)−3(x+5)⇒6x+6=x2+5x−3x−15⇒x2+5x−3x−15−6x−6=0⇒x2−4x−21=0⇒x2−7x+3x−21=0⇒x(x−7)+3(x−7)=0⇒(x−7)(x+3)=0⇒(x−7)=0 or (x+3)=0⇒x=7 or x=−3\Rightarrow \dfrac{1}{x - 3} + \dfrac{1}{x + 5} = \dfrac{1}{3}\\[1em] \Rightarrow \dfrac{(x + 5) + (x - 3)}{(x - 3)(x + 5)} = \dfrac{1}{3}\\[1em] \Rightarrow \dfrac{2x + 2}{(x - 3)(x + 5)} = \dfrac{1}{3}\\[1em] \Rightarrow 3(2x + 2) = (x - 3)(x + 5)\\[1em] \Rightarrow 6x + 6 = x(x + 5) - 3(x + 5)\\[1em] \Rightarrow 6x + 6 = x^2 + 5x - 3x - 15\\[1em] \Rightarrow x^2 + 5x - 3x - 15 - 6x - 6 = 0\\[1em] \Rightarrow x^2 - 4x - 21 = 0\\[1em] \Rightarrow x^2 - 7x + 3x - 21 = 0\\[1em] \Rightarrow x(x - 7) + 3(x - 7) = 0\\[1em] \Rightarrow (x - 7)(x + 3) = 0\\[1em] \Rightarrow (x - 7) = 0 \text{ or } (x + 3) = 0\\[1em] \Rightarrow x = 7 \text{ or } x = -3\\[1em]

Since, age cannot be negative.

Hence, the present age of Joseph = 7 years.

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