One number is equal to the square of another. Find the numbers if their sum is 72.?

Science & Mathematics by Anonymous 2018-07-17 18:33:19

Social Science

10 answers

Anonymous

x = y² y² + y = 72 y² + y - 72 = 0 [ y + 9 ] [ y - 8 ] = 0 y = - 9 , y = 8 x = 81 , x = 64 Numbers are 8 and 64 or - 9 and 81 y = 6
Anonymous

x^2 = y = 72 – x https://www.wolframalpha.com/input/?i=y+... (x – 8)(x + 9) = 0, solutions are (8, 64), (-9, 81)
Anonymous

8,64
Anonymous

x = y^2 x + y = 72 y^2 + y - 72 = 0 (y + 9)(y - 8) = 0 Solutions: x = 64, y = 8 x = 81, y = -9
Anonymous

x^2 + x = 72 x^2 + x - 72 = 0 (x + 9)(x - 8) = 0 x = -9 or x = 8 The numbers are {-9, 81} or {8, 64}.
Anonymous

x + y = 72 x² = y x² + x - 72 = 0 (x+9)(x-8) = 0 x = -9, 8 Two solutions: -9 and 81 or 8 and 64
Anonymous

x+y = 72 y = 72-x y = x² x² = 72-x x² + x - 72 = 0 x = -9,8 (x,y) = (-9, 81) or (8,64)
Anonymous

x^2 + x = 72 x^2 + x - 72 = 0 (x+9)(x-8) = 0 x = -9, x^2 = 81, sum is 72 or x = 8, x^2 = 54, sum is 72
Anonymous

n + n² = 72 n² + n - 72 = 0 (n + 9)(n - 8) = 0 The numbers are either 8 and 64, or -9 and 81.
Anonymous

8