The length of a rectangle is 3 more than twice the width. The area of the rectangle is 119 square inches. What are the dimensions of the rectangle?
width --- x
length --- 2x + 3
area = length*width
= x(2x+3)
x(2x+3) = 119
2x^2 + 3x - 119 = 0
solve using your favourite method of solving quadratics
Let's start by assigning variables to the width and length of the rectangle.
Let's say the width of the rectangle is x inches.
According to the problem, the length of the rectangle is 3 more than twice the width.
So, the length = 2x + 3.
The area of a rectangle is given by the formula: A = length x width.
Therefore, we have the equation: A = (2x + 3) x x.
Since the area is given as 119 square inches, we can substitute this into the equation: 119 = (2x + 3) x x.
To solve this equation, we'll simplify and solve for x.
119 = (2x^2 + 3x).
Rearranging the equation, we get: 2x^2 + 3x - 119 = 0.
Now we can solve this quadratic equation using factoring, completing the square, or using the quadratic formula.
Factoring gives us: (2x - 7)(x + 17) = 0.
This gives us two possible solutions: 2x - 7 = 0 or x + 17 = 0.
Solving the first equation, we find: 2x = 7, x = 7/2, x = 3.5.
Solving the second equation, we find: x = -17.
Since a negative length is not meaningful in this context, we discard the solution x = -17.
Therefore, the width of the rectangle is 3.5 inches.
To find the length, we substitute this value back into the equation: length = 2x + 3.
Therefore, the length of the rectangle is: length = 2(3.5) + 3 = 7 + 3 = 10 inches.
So, the dimensions of the rectangle are width = 3.5 inches and length = 10 inches.
To find the dimensions of the rectangle, we can set up two equations based on the given information:
Let's assume the width of the rectangle is "w" inches.
According to the problem, the length of the rectangle is 3 more than twice the width, so the length can be represented as (2w + 3) inches.
Now, we can set up the first equation:
Area = Length * Width
119 = (2w + 3) * w
Expanding the equation, we get:
119 = 2w^2 + 3w
Rearranging the equation to simplify, we have:
2w^2 + 3w - 119 = 0
Now, we can solve this quadratic equation to find the value(s) of "w" and then substitute it back into the equations to find the dimensions.
To solve the quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
w = (-b ± √(b^2 - 4ac)) / (2a)
Using the coefficients from our equation:
a = 2, b = 3, c = -119
Substituting the values into the formula, we get:
w = (-3 ± √(3^2 - 4 * 2 * -119)) / (2 * 2)
Calculating within the square root:
w = (-3 ± √(9 + 952)) / 4
w = (-3 ± √961) / 4
w = (-3 ± 31) / 4
Now, we have two possible values for the width, which are:
w₁ = (-3 + 31) / 4 = 7
w₂ = (-3 - 31) / 4 = -8
Since dimensions cannot be negative, we can discard the negative value.
So, the width of the rectangle is 7 inches.
Now, we can substitute this value back into one of the initial equations to find the length:
Length = 2w + 3
Length = 2 * 7 + 3
Length = 14 + 3
Length = 17
Therefore, the dimensions of the rectangle are:
Width = 7 inches
Length = 17 inches.