the length of rectangle is 5 cm more than twice its width. if the area is 52 cm^2,find its dimensions

2x² + 5x - 52 = 0

(2x + 13) (x - 4) = 0

x = 4

Can you solve the quadratic:

x(2x + 5) = 52 ?

To find the dimensions of the rectangle, we first need to set up equations based on the given information. Let's assume the width of the rectangle is "w" cm.

According to the problem, the length of the rectangle is 5 cm more than twice its width. Therefore, the length can be expressed as (2w + 5) cm.

The formula for the area of a rectangle is length multiplied by width. We are given that the area is 52 cm^2. So we can set up the equation as follows:

Area = Length × Width
52 cm^2 = (2w + 5) cm × w cm

To solve this equation, we can multiply the terms on the right-hand side:

52 cm^2 = 2w^2 + 5w cm^2

Now, rearrange the equation to form a quadratic equation:

2w^2 + 5w - 52 = 0

To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:

w = (-b ± √(b^2 - 4ac)) / 2a

Here, a = 2, b = 5, and c = -52.

Now, substitute the values of a, b, and c into the formula and solve for w:

w = (-5 ± √(5^2 - 4 * 2 * -52)) / (2 * 2)
w = (-5 ± √(25 + 416)) / 4
w = (-5 ± √441) / 4
w = (-5 ± 21) / 4

So, we have two possible values for the width:
w1 = (-5 + 21) / 4 = 4 cm
w2 = (-5 - 21) / 4 = -6 cm (Discard this negative value since width cannot be negative in this context)

Now, substitute the value of w = 4 cm back into the equation for the length:

Length = 2w + 5 = 2 * 4 + 5 = 8 + 5 = 13 cm

Therefore, the dimensions of the rectangle are 4 cm (width) and 13 cm (length).