Question

The sum of two numbers is 9 and the sum of their squares is 41. What is the larger number?

Answer 1

5 because 5+4 = 9 and 5^2+4^2= 41
Hope this helps!

Answer 2

The larger number is 5.

Explanation

Lets assume, the larger number is $a$ and the smaller number is $b$

As the sum of two numbers is 9, so...

$a+b= 9 ...............................(1)$

Now, the sum of their squares is 41, so....

$a^2 + b^2 = 41 .......................................(2)$

First, solving equation (1) for $b$....

$b=9-a$

Now, plugging this $b=9-a$ into equation (2) , we will get...

$a^2 +(9-a)^2 = 41\\ \\ a^2 +81-18a+a^2 =41 \\ \\ 2a^2-18a+81 =41 \\ \\ 2a^2-18a+40=0\\ \\ 2(a^2 -9a+20)=0\\ \\ a^2 -9a+20=0\\ \\ (a-5)(a-4)=0\\ \\ a=5,4$

If $a=5 ,$ then $b=9-5=4$

So, the larger number is 5.