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

Area of rectangle = $l e n g t h \times b r e a d t h$

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

Let the breadth of the field be $x$ .

Length of the field= $2 x + 1$

Area of the rectangular field= $x (2 x + 1) = 3$

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

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

We get the equation $2 x^{2} + x - 3 = 0$ .

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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:

The correct option is C 2x2+x−3=0 Area of rectangle = length×breadth Given, length =(2×breadth+1) Let the breadth of the field be x. Length of the field= 2x+1 Area of the rectangular field= x(2x+1)=3 2x2+x=3 2x2+x−3=0 We get the equation 2x2+x−3=0.