For a rectangular field of area 3 sq. units, the length is one unit more than twice the breadth $‘ x^{'}$ . The quadratic equation representing the situation is:

$x^{2}$ – $2 x + 6 = 0$

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$x^{2}$ – $2 x + 3 = 0$

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$2 x^{2} + x - 3 = 0$

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$2 x^{2} + x - 6 = 0$

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Solution

The correct option is C
$2 x^{2} + x - 3 = 0$

$A r e a o f r e c t a n g l e = l e n g t h \times b r e a d t h$

Given, $l e n g t h = (2 \times b r e a d t h + 1)$

Let the breadth of the field be $x$ .

Length of the field= $2 x + 1$

Area of the rectangular field

$= x (2 x + 1) = 3$

$= 2 x^{2} + x = 3$

$= 2 x^{2} + x - 3 = 0$

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