# 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).

• 33

x*y=56

x2+y2+x+y=128

x2+y2+2xy+x+y=128+112

(x+y)2+x+y=240

(x+y)square=225

x+y=15

therefore x=8 and y=7

hope this helps u

• -8
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