Answer:
Let the integers be x and y
The sum of the squares of two positive integers is 193.
[tex]x^{2} +y^{2} =193[/tex]
The product of the two integers is 84.
[tex]xy=84[/tex] or
[tex]2xy=168[/tex] ......(2)
Adding both equations (1) and (2)
[tex]x^{2} +y^{2}+2xy =361[/tex]
So, we get [tex](x+y)^{2} =361[/tex]
[tex]\sqrt{(x+y)^{2}} =\sqrt{361}[/tex]
[tex]x+y=19[/tex]
So, the sum of the two positive integers is 19.
Now we get; [tex]x=19-y[/tex]
Substituting in xy=84;
[tex](19-y)y=84[/tex]
=> [tex]19y-y^{2}=84[/tex]
Equating to zero, we get;
=> [tex]y^{2} -19y+84=0[/tex]
=> [tex]y^{2} -12y-7y+84=0[/tex]
=> [tex]y(y-12)-7(y-12)=0[/tex]
=> [tex](y-7)(y-12)=0[/tex]
Both are positive roots and the answer.
Hence, the two numbers are 7 and 12.
We can check :
[tex](7)^{2}+(12)^{2}=193[/tex]
[tex]49+144=193[/tex]
[tex]193=193[/tex]
And
[tex]7\times12=84[/tex]
And sum =[tex]7+12=19[/tex]
