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:

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Solution

The correct option is C

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

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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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 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

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: