Question

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

• A. $6,10$
• B. $6,4$
• C. $4,10$
• D. None of these

Answer 1

Answer: (A)

$6,10$

Let the larger part be $x$

So, smaller part is $16-x$

Given,

$2\times x^2={(16-x)}^2+164$

$2x^2=(256-32x+x^2)+164$

$x^2+32x-420=0$

$x^2-10x+42x-420=0$

$x(x-10)+42(x-10)=0$

$(x-10)(x+42)$

$\to x=10,-42$

$\because x$ cannot be negative, hence $x=10$

So, the two parts are $10,6$