The length of a rectangle is 2 more than twice its width. If the area of the rectangle is 24 sq. m., find its dimensions.
L = 2 + 2 W = 2 ( 1 + W )
A = L * W
24 = 2 * ( 1 + W ) * W
24 = 2 * ( W + W ^ 2 ) Divide both sides by 2
12 = W + W ^ 2 Subtract 12 to both sides
0 = W + W ^ 2 - 12
Now you must solwe equation :
W ^ 2 + W - 12 = 0
The exact solutions are 3 and - 4
Width cannot be negative so W = 3 m
L = 2 + 2 W
L = 2 + 2 * 3
L = 2 + 6 = 8 m
A = L * W = 8 * 3 = 24 m ^ 2
P. S.
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W ^ 2 + W - 12 = 0
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W ^ 2 + W - 12 = 0
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To find the dimensions of the rectangle, we need to set up an equation using the given information.
Let's represent the width of the rectangle as "w" and the length as "l".
According to the problem, the length of the rectangle is 2 more than twice its width. So we can write this as:
l = 2w + 2
The area of a rectangle is given by the formula: Area = length * width.
We are given that the area of the rectangle is 24 sq. m. So we can write this as:
24 = l * w
Now, substitute the value of l from the first equation into the second equation:
24 = (2w + 2) * w
Expanding this equation:
24 = 2w^2 + 2w
Rearranging the equation to make it a quadratic equation in standard form:
2w^2 + 2w - 24 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, let's use factoring. Factoring out a common factor of 2:
2(w^2 + w - 12) = 0
Now we have a quadratic equation that can be factored:
(w + 4)(w - 3) = 0
Setting each factor equal to zero:
w + 4 = 0 or w - 3 = 0
Solving these equations for w:
w = -4 or w = 3
Since the width of a rectangle cannot be negative, we take w = 3.
Now substitute the value of w back into the first equation to find the length:
l = 2w + 2
l = 2*3 + 2
l = 6 + 2
l = 8
Therefore, the dimensions of the rectangle are width = 3 meters and length = 8 meters.