The length of a rectangle is 5 meters more than twice the width, and the area of a rectangle is 42 square meters. Find the length and width of the rectangle.
width --- x
length ---- 2x+5
x(2x+5) = 42
2x^2 + 5x - 42 = 0
solve for x
hint: it factors
To find the length and width of the rectangle, we need to use the information provided and solve the problem step by step.
Let's assign variables to represent the length and width of the rectangle. Let L be the length and W be the width.
From the given information, we have two conditions:
1. The length of the rectangle is 5 meters more than twice the width: L = 2W + 5.
2. The area of the rectangle is 42 square meters: LW = 42.
Now we can solve these two equations simultaneously to find the values of L and W.
Substitute the value of L from the first equation into the second equation:
(2W + 5)W = 42
Simplify the equation:
2W² + 5W = 42
Rearrange the equation to make it a quadratic equation:
2W² + 5W - 42 = 0
Now we can solve this quadratic equation. We can factor it or use the quadratic formula.
Factoring:
(2W - 7)(W + 6) = 0
This gives us two possible solutions:
1. 2W - 7 = 0 => W = 7/2
2. W + 6 = 0 => W = -6
Since we are dealing with the dimensions of a rectangle, the width cannot be negative. Therefore, the width W = 7/2 or 3.5 meters.
Now substitute the value of W into the first equation to find the length:
L = 2W + 5 = 2(3.5) + 5 = 7 + 5 = 12 meters
So, the length of the rectangle is 12 meters and the width is 3.5 meters.