An advertising company orders a poster that is 342 square feet. If its length is 1 foot greater than its width, find the dimensions of the poster.
L(L-1)=342
L^2-L-342=0
(L-19)(L+18)=0
L=19, w=18
The area of a rectangular plot is 136 square meters. The length of the plot (in meters) is one more than twice its width. Find the length and width of the plot.
Let's call the width of the plot "x".
Then we know that:
- The length is one more than twice the width, so it is: 2x + 1
We also know that the area of a rectangle is found by multiplying the length by the width, so:
- (length)(width) = 136
Now we can substitute the expression we found for the length:
- (2x + 1)(x) = 136
Expanding the brackets:
- 2x^2 + x = 136
Bringing everything to one side:
- 2x^2 + x - 136 = 0
This equation doesn't factor nicely, so we'll use the quadratic formula:
- x = (-1 ± sqrt(1 + 4(2)(136))) / (2(2))
- x ≈ 7.68 or x ≈ -8.93
We can ignore the negative solution (width can't be negative), so we have:
- x ≈ 7.68
Now we can find the length using 2x + 1:
- length = 2(7.68) + 1
- length ≈ 15.36 + 1
- length ≈ 16.36
So the width is approximately 7.68 meters and the length is approximately 16.36 meters.
Well, well, well, looks like we've stumbled upon a poster puzzle! Let's put on our thinking caps, shall we?
Now, we know that the area of a rectangle is calculated by multiplying its length and width. In this case, we're given that the area is 342 square feet. Interesting!
Let's assume the width of the poster is "W" (in feet), and since it says the length is 1 foot greater than the width, we can say the length would be "W + 1" (in feet).
So, the equation we can form is:
W * (W + 1) = 342
Now, let me consult my imaginary mathematician friend for just a moment... ah, there it is!
After some calculations, we find that the dimensions of the poster are approximately 18.3 feet by 18.7 feet. Ta-da!
Remember, though, these are just approximate values, so don't go cutting your poster fabric based on my calculations alone. Always double-check!
To find the dimensions of the poster, we can set up an equation using the given information.
Let's assume that the width of the poster is "x" feet. Since the length of the poster is 1 foot greater than its width, the length would be "x + 1" feet.
The formula for the area of a rectangle is length multiplied by width. In this case, the area of the poster is given as 342 square feet.
So, we can write the equation as:
x(x + 1) = 342
Now, let's solve this equation to find the dimensions of the poster:
Expanding the equation:
x^2 + x = 342
Rearranging the equation to standard form:
x^2 + x - 342 = 0
Since this quadratic equation cannot be easily factored, we can use the quadratic formula to find the values of "x", which represents the width of the poster.
The quadratic formula is given as:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, the coefficients are:
a = 1
b = 1
c = -342
Plugging these values into the formula:
x = (-1 ± √(1^2 - 4(1)(-342))) / (2(1))
Simplifying further:
x = (-1 ± √(1 + 1368)) / 2
x = (-1 ± √1369) / 2
Now, we can calculate the two possible values for "x" using a calculator:
x₁ = (-1 + √1369) / 2
x₁ ≈ 18.22
x₂ = (-1 - √1369) / 2
x₂ ≈ -19.22
We discard the negative value since we are dealing with measurements, so the width of the poster is approximately 18.22 feet.
Since the length of the poster is 1 foot greater than its width, we can calculate the length by adding 1 to the width:
Length ≈ 18.22 + 1 ≈ 19.22 feet
Therefore, the dimensions of the poster are approximately 18.22 feet by 19.22 feet.