Question

If the sum of two numbers is $9$ and the sum of their squares is $41$, then the numbers are

• A. $4$ and $5$
• B. $1$ and $8$
• C. $3$ and $6$
• D. $2$ and $7$

Answer 1

The correct option is A $4$ and $5$

Let one of the numbers be $x$. The the other no. will be $(9-x)$
$\therefore x^2+(9-x{)}^2=41$
$x^2+81-18x+x^2=41$
$2x^2-18x+40=0$
$x^2-9x+20=0$
$(x-5)(x-4)=0$
$\therefore x=5$ or $x=4$
Then $y=9-x=9-5=4$
or $y=9-4=5$

Hence, option A.