you are making a rectangular table. The area of the table should be 10 ft^2. You want the length of the table to be 1 ft shorter than twice its width. What should the dimensions of the table be?
This question has to be quadratic equation!
I really need to know the answer, because I have a test tomorrow. THANK YOU
just solve it
To solve this problem, we can set up a quadratic equation using the given information.
Let's assume the width of the table is 'x' ft. According to the problem, the length of the table should be 1 ft shorter than twice its width. Therefore, the length would be (2x - 1) ft.
We know that the area of a rectangular table is length multiplied by width. In this case, the area is given as 10 ft^2.
Using the formula for the area, we can set up the equation:
Area = Length * Width
10 = (2x - 1) * x
Now, we need to simplify and solve this quadratic equation to find the value of 'x.'
First, expand the equation:
10 = 2x^2 - x
Next, move all terms to one side to set the equation to zero:
2x^2 - x - 10 = 0
This is a quadratic equation in standard form. Now, we can either factor or use the quadratic formula to solve for 'x.'
To factor this quadratic equation, we need to find two numbers that multiply to give -20 (product of the coefficient of x^2 and the constant term) and add up to give -1 (coefficient of x).
After trying different combinations, we find that the factors are -5 and 4:
(2x - 5)(x + 2) = 0
Setting each factor to zero, we get:
2x - 5 = 0 or x + 2 = 0
Solving for 'x' in each equation results in:
2x = 5 or x = -2
x = 5/2 or x = -2
Since width cannot be negative (as it represents a physical length), we discard the negative solution.
Therefore, the width of the table is 5/2 ft or 2.5 ft.
To find the length, we substitute this value back into the expression for the length: 2x - 1.
Length = 2(5/2) - 1 = 5 - 1 = 4 ft
So, the dimensions of the table should be 2.5 ft by 4 ft.
LW = Area
L = 2W - 1
Substitute 2W-1 for L.
(2W-1)W = 10
Work it from there.