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the length of a rectangle is 3 meters less than twice the width. If the area of the rectangle is 537 square​ meters, find the dimensions

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Let's call the width "x"

Twice the width would be 2x, and 3 meters less than that would be 2x-3.

Area of a rectangle is length times width, so (2x-3) times x.

We know the area is 665, so we can model this information with x(2x-3)=665...which is a quadratic equation.

To solve this equation, start by distributing the x to both the 2x term and the -3 term. Now we have 2x2-3x=665.

There are several methods for solving a quadratic equation, but since this is not easily factorable I will use the quadratic formula. This formula gives the solutions to a quadratic expression that is equal to 0, so we will have to subtract 665 from each side to set that up.

Now 2x2-3x-665=0, giving us a=2, b=-3, and c=-665.

Substituting these values into the quadratic formula we have x=-(-3)±√(9–4·2·-665) all divided by 2·2

Simplifying the parts: -(-3) equals 3, (9–4·2·-665) equals 5329 and its square root is 73, and the denominator 2·2 equals 4.

This gives us (3±73)/4.

Evaluating the "plus version" 3+73=76 and 76/4=19

Evaluating the "minus version" 3-73=-70 and -70/4=-17.5

Referring back to the original question: x is the width of the rectangle, so it cannot be a negative number. That means the width must be 19. The length was 3 less than twice the width (or 2·19-3), which is 35. The dimensions of the rectangle are: length=35 meters and width=19 meters. (We can check this by multiplying length times width, 35·19=665. Since 665 was the given area, we must be correct!)

User Ben Thielker
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