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QUESTION: Here is a link :

https://archive.org/stream/intermediatealge00slaurich#page/n203/mode/2up

Thanks again.

ANSWER: I had some problems understanding this question, because

"width" and "length" were used interchangeably

and they omitted important details!

the shaded areas within a 6" square are congruent

the unshaded areas are square

(only one of these facts is needed to derive the other)

Just consider the right picture for now. The shaded areas are congruent rectangles and the white areas are squares.

What are the dimensions of the shaded rectangles?

Let x be the longer dimension of a shaded rectangle. Call it the length.

Since the large rectangle is 6 inches on a side and the shaded rectangles are congruent, the width of a shaded rectangle must be (6-x)/2 inches.

The area of a shaded rectangle is x(6-x)/2.

The total shaded area is 4x(6-x)/2 = 2x(6-x) = 12x-2x².

Since the total shaded area is 4/9 of the whole square,

12x-2x² = (4/9)36 = 16 in²

2x²-12x+16 = 0

By the quadratic formula,

x = [12 ± √(12² – 4·2·16)] / [2·2]

= [12 ± √16] / 4

= [12 ± 4] /4

= 4, 2

Note that there are two solutions for x.

If x = 4, the rectangles are 4 inches by (6-4)/2 inches, or 4 inches by 1 inch. That is the picture on the right.

If x = 2, the rectangles are 2 inches by (6-2)/2 inches, or 2 inches square. That is the picture on the left.

The point of the question was to show that a quadratic equation can have two different solutions. (It doesn't always have two solutions, and sometimes one of the solutions is not valid in the context of the question.)

---------- FOLLOW-UP ----------

QUESTION: " the width of a shaded rectangle must be (6-x)/2 inches.

The area of a shaded rectangle is x(6-x)/2.

The total shaded area is 4x(6-x)/2 = 2x(6-x) = 12x-2x²."

I am not clear on this part of your solution. Thanks again.

Note: Sorry for the repost of the link.

The shade rectangles are congruent, so the white rectangles in the corner are square and congruent.

https://www.flickr.com/photos/dwread/15635269537/

Consider the top row, which contains two white squares and a shaded rectangle.

If the length of the rectangle is x, that leaves 6-x inches for both white squares.

Since the white squares are congruent, each square is (6-x)/2 inches square.

The width of the shaded rectangle is the same as the length of a white square, or (6-x)/2 inches.

area of the shaded rectangle = length · width = x·(6-x)/2 in²

There are four shaded rectangles, for a total of 4·x·(6-x)/2 = 2x(6-x) = 12x-2x² in².

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