SOLUTION: QUADRATIC EQUATION 1) THE AREA OF A RECTANGULAR GROUND IS 54SQUARE M. IF THE LENGTH OF THE GROUND IS 2M LESS AND 1M MORE, IT WILL HAVE BEEN A SQUARE. FIND LENGTH AND BREADTH OF TH

Question 1084371: QUADRATIC EQUATION
1) THE AREA OF A RECTANGULAR GROUND IS 54SQUARE M. IF THE LENGTH OF THE GROUND IS 2M LESS AND 1M MORE, IT WILL HAVE BEEN A SQUARE. FIND LENGTH AND BREADTH OF THE GROUND
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the area of the rectangle is 54 square meters.

if the length is 2 meters less and the width is 1 meter more, it will have been a square.

assume the length of the rectangle is x and the width of the rectangle is y.

in order for it to be a square, you need to shorten the length by 2 meters and increase the width by 1 meter.

you would get x-2 is the length and y + 1 is the width.

if it is now a square, then x-2 = y+1 because all sides are equal in a square.

you can solve for y to get y = x-3

go back to your rectangle.

the area is 54 square meters which is equal to x * y.

since y is equal to x-3, you get x * (x-3) = 54

simplify to get x^2 - 3x = 54

subtract 54 from both sides to get x^2 - 3x - 54 = 0

factor to get (x-9) * (x+6) = 0

solve for x to get x = 9 or x = -6

x can't be negative so x has to be 9

since x * y = 54, when x = 9, y has to be equal to 6

you get x * y = 9 * 6 = 54

it's a rectangle.

if you subtract 2 from the length and add 1 to the width, you get:

(9-2) * (6+1) = 7 * 7 = 49

it's now a square with each side equal to 7 and the area of the square is 49 square meters.

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