The product of two numbers is 56. When their sum is added to the sum of their squares, we get 128. Find the numbers.
Let the two numbers be x and y.
Given, xy = 56
... (1)
According to the given condition,
(x + y) + (x2 + y2) = 128
⇒ (x + y) + (x2 + y2 + 2xy) - 2xy = 128
⇒ (x + y) + (x + y)2 - 2 × 56 = 128
⇒ (x + y)2 + (x + y) = 128 + 112
⇒ (x + y)2 + (x + y) = 240
Now, let x + y = z
Therefore, z2 + z = 240
⇒ z2 + z - 240 = 0
⇒ z2 + 16z -15z - 240 = 0
⇒ z (z + 16) -15(z + 16) = 0
⇒ (z -15) (z + 16) = 0
⇒ z = 15, -16
Now, x + y ≠ -16, because on solving it with respect to (1), we get irrational roots.
Hence, x + y = 15
⇒ x + = 15 [using (1)]
⇒ x2 -15x + 56 = 0
⇒ x2 -8x -7x + 56 = 0
⇒ (x-8)(x-7) = 0
⇒ x = 8, 7
On putting value of x = 8, 7 in (1) we get y = 7, 8
Hence, the two number are (8, 7) or (7, 8).