Solution. Because , = 2· 1,00010,000 +1. Similarly, X = 3· 10,000100,000 + 1. Therefore, Y ≈ Compute the number of quadratic functions = + + with integer roots and integer f(x) ax2 bx c coefficients whose graphs pass through the points (0, 0) and (15, 225). Solution. Because the passes through (0, 0), conclude that = 0 and = + c f(x) ax2 bx. Then, = 225 + = 225a + 15b = 225, from which = 15 15a. f(15) a(15)2 b(15) b − ⇒ On the other hand, can be factored as = + so if the roots are integers, f f(x) ax(x b/a), b/a must be an integer. Divide both sides of the equation = 15 15a by to obtain b a − = 15/a Thus 15/a must be an integer, and {±1,±3,±5,±15}. b/a −15. a Because = 15 15a is linear, each of these values for yields a unique integer value for b a b, − so there are 8 such ordered pairs. FYI. The values of and the a, b, nonnegative root are given in the table:
Solution. Because , X = 2· 1,00010,000 +1. Similarly, Y = 3· 10,000100,000 + 1. Therefore, ≈ Compute the number of quadratic functions f(x) = ax2 + bx + c with integer roots and integer coefficients whose graphs pass through the points (0, 0) and (15, 225). Solution. Because the passes through (0, 0), conclude that c = 0 and f(x) = ax2 + bx. Then, f(15) = 225 ⇒ a(15)2 + b(15) = 225a + 15b = 225, from which b = 15 − 15a. On the other hand, f can be factored as f(x) = ax(x + b/a), so if the roots are integers, b/a must be an integer. Divide both sides of the equation b = 15 − 15a by a to obtain b/a = 15/a −15. Thus 15/a must be an integer, and a {±1,±3,±5,±15}. Because b = 15 − 15a is linear, each of these values for a yields a unique integer value for b, so there are 8 such ordered pairs. FYI. The values of a, b, and the nonnegative root are given in the table: