Question

Divide 16 into two parts such that the twice the square of the larger part exceeds the square of smaller part by 164.

Answer 1

HOPE THIS HELPS YOU!! :D

Answer 2

Answer:

6 and 10

Step-by-step explanation:

let the two parts of 16 be x and y. (x larger and y smaller)

therefore x + y = 16

therefore, x = 16 - y  (I)

it is given that twice the square of the larger part exceeds the square of the smaller part by 164,

therefore, 2x^2 - y^2 = 164  (II)

substituting the value of x from equation (i) in equation (II)

therefore, 2( 16 - y)^2 - y^2 = 164

2(256 - 32y + y^2) - y^2 = 164        [ (a-b)^2 = a^2 - 2ab - b^2]

512 - 64y + 2y^2 - y^2 = 164

y^2 - 64y + 348 = 0                      [512 - 164 = 348]

y^2 - 58y - 6y + 348 =0  

y=58  or y=6  [ factorization method]

since 58 is bigger than 16 it is not acceptable

therefore y=6

x = 16 - y

therefore x = 16 - 6

x = 10

Hope this helps.