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

Hint: We are given to divide the given number into such parts such that the twice the square of the larger part exceeds the square of the smaller parts by $ 164 $ . We can solve such questions by making an equation and then solving it. The equation will come out to be quadratic but we will solve it in one variable. The second variable will be related to the number $ 16 $ .

Complete step-by-step answer:
We will first make two equations as there are two numbers to be found, but one equation which is quadratic should only contain one variable,
Let the larger number be $ x $ ,and the smaller number be $ y $ ,
 $ 16 - x = y $
Also we have to Divide $ 16 $ into two such parts such that the twice the square of the larger part exceeds the square of the smaller parts by $ 164 $ , so we write,
 $ 2{x^2} - {(16 - x)^2} = 164 $
The equation upon further simplification yields,
\[{x^2} + 32x - 420 = 0\]
The quadratic equation will be solve by factorization method,
\[(x + 42)(x - 10) = 0\]
Which gives
\[
  x = - 42\;,or \\
  x = 10 \;
\]
Thus the value of $ x $ is $ 10 $ and the second number is ,
 $ 16 - 10 = 6 $ ,
Thus the number $ 16 $ can be divided as,
 $ 10{\text{ and 6}} $ .

Note: The question’s quadratic equation given above was solved by the help of the factorization but if it was not solvable this way then the quadratic equation formula could have been deployed to solve it, the method is,
To solve equation of the form,
 $ a{x^2} + bx + c = 0 $ , we get $ x $ as
 $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $