Hence, the two integers could be:
3 and 9
0R
-3 and -9
Let the two integers be denoted by a and b.
It is given that:
One integer is 1/3 times another.
i.e. let
[tex]a=\dfrac{1}{3}b[/tex]
i.e.
[tex]b=3a[/tex]
Also,
The sum of their squares is 90.
i.e.
[tex]a^2+b^2=90\\\\i.e.\\\\a^2+(3a)^2=90\\\\i.e.\\\\a^2+9a^2=90\\\\i.e.\\\\10a^2=90\\\\i.e.\\\\a^2=9\\\\i.e.\\\\a=\pm 3[/tex]
Hence, when a=3 we have:
[tex]b=9[/tex]
and when,
[tex]a=-3[/tex]
then b=-9