I don't get this at all

A rectangle is 9 feet longer than it is wide, and its area is 36 square feet. Find its dimensions.

can someone help me please :D :x

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I don't get this at all

A rectangle is 9 feet longer than it is wide, and its area is 36 square feet. Find its dimensions.

can someone help me please :D :x

A rectangle is 9 feet longer than it is wide, and its area is 36 square feet. Find its dimensions.

can someone help me please :D :x

- Laurel

Laurel wrote:I don't get this at all

A rectangle is 9 feet longer than it is wide, and its area is 36 square feet. Find its dimensions.

can someone help me please :D :x

The above is a word equation. You have to reduce it to a math equation representing the information, but solvable.

The area of a rectangle is length x width. You don't know the length and width but you know something about them. We will assign a symbol (the letter x is customary, but I'll use y to avoid confusion with multiplicayion). I don't know the width, but I'll call it y. The length is 9' longer or y+9.

The area is y*(y+9) =36

This can be rewritten as y^2 + 9*y -36 = 0

This can be solved by the quadratic formula, or you may see how to factor it. In fact, it factors

(y+12)*(y-3) =0

There are two solutions, when each of the factors is zero. y= 3 is one solution (the other is y = -12, which makes no physical sense and is discarded. It solves the math, but not the real world problem). If the width is 3, the length is 12. As a check, 3*12 = 36.

- Guest

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