Question

The sum of two numbers is $9$. The sum of their reciprocals is 1/2. Find the sum of squares of the numbers.

Answer 1

$\Rightarrow$ Let one number be $x$, then other number will be $9-x$

According to the question,

$\Rightarrow$ $\frac{1}{x}+\frac{1}{9-x}=\frac{1}{2}$

$\Rightarrow$ $\frac{9-x+x}{x(9-x)}=\frac{1}{2}$

$\Rightarrow$ $\frac{9}{9x-x^2}=\frac{1}{2}$

$\Rightarrow$ $2\left(9\right)=9x-x^2$

$\Rightarrow$ $18-=9x-x^2$

$\Rightarrow$ $x^2-9x+18=0$

$\Rightarrow$ $x^2-6x-3x+18=0$

$\Rightarrow$ $x(x-6)-3(x-6)=0$$

$\Rightarrow$ $(x-6)(x-3)=0$

$\Rightarrow$ $x-6=0$ and $x-3=0$

$\therefore$ $x=6$ and $x=3$

$\Rightarrow$ One number is $6$ and other number is $3$

$\Rightarrow$ The sum of squares of the numbers $=(6{)}^2+(3{)}^2=36+9=45$