Question

Divide 16 in two parts such that twice square of larger part exceed the smaller part by 164

Answer 1

Let the larger part is x
Smaller part is 16-x
2xv2 is greater than( 16-x)+164

X is greater than or equal to 10

Answer 2

Let the two required numbers be x and y such that x > y.

Then,

$\sf\green{x + y = 16 \: and \: {2x}^{2} - {y}^{2} = 164}$

$\longrightarrow$$\sf\red{ {2x}^{2} - {(16 - x) }^{2} = 164}$

$\longrightarrow$$\sf\red{{x}^{2} + 32x - 420 = 0}$

$\longrightarrow$$\sf\red{(x -1 0)(x + 42) = 0}$

$\longrightarrow$$\sf\red{x = 10 \: or \: x = - 42}$

But x ≠ -42 ,so x = 10

Hence, the required two numbers are 10 & 6.