Five times of a positive integer is 3 less than twice the square of that number.

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Question

Five times of a positive integer is 3 less than twice the square of that number. The number is -

Options:

A . 3

B . 13

C . 23

D . 33

Answer: Option A
$$\eqalign{
& {\text{Let }}x{\text{ is the number}} \cr
& {\text{According to question,}} \cr
& 5x = 2{x^2} - 3 \cr
& 2{x^2} - 5x - 3 = 0 \cr
& 2{x^2} - 6x + x - 3 = 0 \cr
& 2x\left( {x - 3} \right) + 1\left( {x - 3} \right) = 0 \cr
& \left( {2x + 1} \right)\left( {x - 3} \right) = 0 \cr
& x = 3 \cr} $$

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