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 the larger part be$x.$

Then, the smaller part$=16-x$

By hypothesis, we have $2x^2={(16-x)}^2+164$

$$\begin{gathered} \Rightarrow x^2+32x-420=0 \\ \Rightarrow (x+42)(x-10)=0 \\ \Rightarrow x=-42orx=10 \end{gathered}$$

∵$x.$can not be a negative

$\Rightarrow x=10$

The larger part $x=10$

The smaller part $16-x=16-10=6$

Hence, the required parts are$\mathbf{10}\mathbf{}$and$\mathbf{}\mathbf{6}$