## Sum of Two Numbers

Two different positive numbers $a$ and $b$ each differ from their reciprocals by $1$. What is $a+b$ ?
Source: NCTM Mathematics Teacher

SOLUTION
We have
$\displaystyle a-\frac{1}{a}=1$
Simplify
$a^2-1=a$
$a^2-a-1=0$
Solve for $a$ by completing the square
$a^2-a+\displaystyle\frac{1}{4}=\frac{5}{4}$
$\displaystyle\left (a-\frac{1}{2}\right )^2=\frac{5}{4}$
$\displaystyle a-\frac{1}{2}=\pm\frac{\sqrt{5}}{2}$
$a=\displaystyle\frac{1\pm\sqrt{5}}{2}$
$=\displaystyle\frac{1+\sqrt{5}}{2}>0$
Similarly
$\displaystyle\frac{1}{b}-b=1$
Simplify
$1-b^2=b$
$b^2+b-1=0$
Solve for $b$ by completing the square
$b^2+b+\displaystyle\frac{1}{4}=\frac{5}{4}$
$\displaystyle\left (b+\frac{1}{2}\right )^2=\frac{5}{4}$
$\displaystyle b+\frac{1}{2}=\pm\frac{\sqrt{5}}{2}$
$b=\displaystyle\frac{-1\pm\sqrt{5}}{2}$
$=\displaystyle\frac{-1+\sqrt{5}}{2}>0$
$a+b=\displaystyle\frac{1+\sqrt{5}}{2}+\frac{-1+\sqrt{5}}{2}$
$=\sqrt{5}$

Answer: $\sqrt{5}$

Advertisements

## About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , . Bookmark the permalink.