Question

One number is six more than another number. The sum of their squares is 90.

Answer 1

One number is six more than the other, so the numbers are x and x+6.

The sum of their squares is 90.
$x^2+(x+6)^2=90 \\ x^2+x^2+12x+36=90 \\ 2x^2+12x+36=90 \ \ \ \ \ \ \ \ \ \ \ |-90 \\ 2x^2+12x-54=0 \ \ \ \ \ \ \ \ \ \ \ \ \ |\div 2 \\ x^2+6x-27=0 \\ x^2+9x-3x-27=0 \\ x(x+9)-3(x+9)=0 \\ (x-3)(x+9)=0 \\ x-3=0 \ \lor \ x+9=0 \\ x=3 \ \lor \ x=-9 \\ \\ \hbox{for x=3: } \\ x+6=3+6=9 \\ \\ \hbox{for x=-9: } \\ x+6=-9+6=-3$

The numbers are: 3 and 9 or -9 and -3.