find two consecutive odd integers such that the sum of their squares is 10 less than 20 times the smaller integer

Answer 1

where do you come up with these questions? they don't look like textbook or class room type questions.

x^2 + y^2 = 20x -10

x is the smaller integer. x=y-2

(y-2)^2 + y^2 = 20(y-2) -10

y^2 -4y + 4 = 20y -40-10

y^2 -24y = -54

y^2 -24y +54 = 0 use the quadratic formula

y= 24/2 + or - (1/2)sqr(24^2 -4(54)) = 12 + or - (1/2)sqr360 = 12 + or about 19/2 = 12+ or- 9.5=

21.5 or 2.5

x=y-2 = 19.5 or 0.5

x = they're sort of "odd" but not integers or consecutive

No solution. DNE = Does Not Exist. Some problems have no solution or "almost solutions" that don't quite fit.

or maybe there was a typo in the question? Meanwhile, I'd go with DNE