Question

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

Answer 1

Let x and (16 - x) are two parts of 16 where (16 - x) is longer and x is smaller.
According to the question,

2 $\times$ square of longer = square of smaller + 164

⇒ $2\times (16-x{)}^2=x^2+164$

⇒ $2\times (256+x^2-32x)=x^2+164$

⇒ $512+2x^2-64x=x^2+164$

⇒ $x^2-64x+512-164=0$

⇒ $x^2-64x+348=0$

⇒ $x^2-58x-6x+348=0$

⇒ $x(x-58)-6(x-58)=0$

⇒ $(x-6)(x-58)=0$

⇒ $x=6and58$

But $x\ne 58$ because $x<16$

so, $x=6$ and $16-x=10$

Hence, the answer is 6 and 10