Question

Divide 20 into two parts such that the twice the square of the smaller part is 16 less than the square of the larger part ​

Answer 1

Question:

Divide 20 into two parts such that the twice the square of the smaller part is 16 less than the square of the larger part .

Answer:

Smaller part = 8

Larger part = 12

Solution:

Let the smaller part be x , then the larger part will be (20-x) .

Now,

According to the question, twice the square of the smaller part is 16 less than the square of the larger part.

Thus,

=> 2x² = (20-x)² - 16

=> 2x² = 20² - 2•20•x + x² - 16

{ (A-B)² = A² - 2•A•B + B² }

=> 2x² = 400 - 40x + x² - 16

=> 2x² - x² + 40x - 400 + 16

=> x² + 40x - 384 = 0

=> x² + 48x - 8x - 384 = 0

=> x(x+48) - 8(x+48) = 0

=> (x+48)(x-8) = 0

=> x = - 48 or 8

Here ,

x = -48 is rejected value

x = 8 is appropriate value

Hence,

The small part = x = 8

The larger part = 20 - x = 20 - 8 = 12