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Answer :

Let the number of articles produced be \(x\).

Thus, the cost of production of each article can be expressed as \(2x + 3\).

As the total cost of production on that day is 90 Rs.,

\(\Rightarrow x(2x + 3) = 90\)

\(\Rightarrow 2x^2 + 3x - 90 = 0\)

(Splitting * 3x * as * -12x + 15x *)

\(\Rightarrow 2x^2 + -12x +15x - 90 = 0\)

\(\Rightarrow 2x(x - 6) + 15(x - 6) = 0\)

\(\Rightarrow(2x + 15)(x - 6) = 0\)

The roots of this equation are the values of x for which \((2x + 15)(x - 6) = 0 \)

which are,

\(2x + 15 = 0 \) or \(x - 6 = 0\)

Thus, \(x = \frac{-15}{2}\) or \(x = 6\)

Since the number of articles produced cannot be a negative number, \(x = \frac{-15}{2}\) is rejected.

** Thus, the number of articles produced is 6 and the cost of each article is \(2×6 + 3\) = 15Rs.
**

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