One side of the square was increased by 4 dm, and the other was reduced by 6 dm.

One side of the square was increased by 4 dm, and the other was reduced by 6 dm. As a result, a rectangle with an area of 56 dm2 was obtained. Find the side length of the square.

All sides of a square are equal.

Let the side of the square be x dm.

If the length of the side of the square is increased by 4 dm, then the length will be equal to: (x + 4) dm.

If the length of the side of the square is reduced by 6 dm, then the length will be equal to: (x – 6) dm.

So, we have a rectangle with sides: a = (x + 4) dm; in = (x – 6) dm.

Its area is 56 dm2. It is calculated by the formula: S = a x in.

We substitute in the formula the values ​​of the lengths of the sides of the rectangle and we get the equation:

(x + 4) x (x – 6) = 56.

We transform the equation by expanding the brackets:

x2 – 6x + 4x – 24 = 56.

x2 – 2x – 24 – 56 = 0.

x2 – 2x – 80 = 0 is a quadratic equation.

Find the discriminant: D = (-2) 2 – 4 x 1 x (-80) = 4 + 320 = 324.

D is greater than zero, so the equation has 2 roots:

x1 = (- (-2) + √324) / 2 = (2 + 18) / 2 = 20/2 = 10.

x2 = (- (-2) – √324) / 2 = (2 – 18) / 2 = -16 / 2 = -8 – this root does not fit, because x is the length of the side of the square, and its value cannot be expressed as a negative number.

Hence, x = 10 dm.

Answer: the length of the side of the square is 10 dm.