The length of a rectangular field is 2 m greater than three times its width. The area of the field is 1496 m2. What are the dimensions of the field?

Mar 9, 2018

Length and width of the field are $68 \mathmr{and} 22$ meter respectively .

Explanation:

Let the width of the rectangular field is $x$ meter, then the

length of the field is $3 x + 2$ meter.

The area of the field is $A = x \left(3 x + 2\right) = 1496$ sq.m

$\therefore 3 {x}^{2} + 2 x - 1496 = 0$ Comparing with standard quadratic

equation ax^2+bx+c=0; a=3 ,b=2 ,c=-1496

Discriminant  D= b^2-4ac; or D =4+4*3*1496=17956

Quadratic formula: $x = \frac{- b \pm \sqrt{D}}{2 a}$or

$x = \frac{- 2 \pm \sqrt{17956}}{6} = \frac{- 2 \pm 134}{6}$

$\therefore x = \frac{132}{6} = 22 \mathmr{and} x = - \frac{136}{6} \approx - 22.66$ . Width can not

be negative, so $x = 22$ m and $3 x + 2 = 66 + 2 = 68$ m. Hence

length and width of the rectangular field is $68 \mathmr{and} 22$ meter

respectively . [Ans]