What should the width of the border of the yellow roses be?

Edit: Picture showed original flower bed to be 8ft by 12ft

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What should the width of the border of the yellow roses be?

Edit: Picture showed original flower bed to be 8ft by 12ft

Re: 10th grade math question--help me [J-No]
[ In reply to ]

Re: 10th grade math question--help me [J-No]
[ In reply to ]

The way that you explained it, you have 8x12 => 96 sq. ft. of red roses. The total area will be twice that, 192 sq. ft. Which means that the new plot would be 12x16, with a two foot border around the original plot.

Of course, you are going to need to use variables in 10th grade.

Let x be the width of the yellow border (it isn't stated, but the border needs to be equal all around, or there are infinite solutions). Drawing it, you will see 8 basic segments, 2 widths, 2 lengths, 4 corners (you can divvy up the segments differently). The area of the yellow bed is the result of all of those areas added together:

2w*x+2l*x + 4x^2 = 96

40x + 4x^2 = 96

0 = x^2 + 10x - 24

0 = (x+12)(x- 2)

checking each factor, and solving for zero, {x = -12} isn't a real answer, so x = 2.

Of course, you are going to need to use variables in 10th grade.

Let x be the width of the yellow border (it isn't stated, but the border needs to be equal all around, or there are infinite solutions). Drawing it, you will see 8 basic segments, 2 widths, 2 lengths, 4 corners (you can divvy up the segments differently). The area of the yellow bed is the result of all of those areas added together:

2w*x+2l*x + 4x^2 = 96

40x + 4x^2 = 96

0 = x^2 + 10x - 24

0 = (x+12)(x- 2)

checking each factor, and solving for zero, {x = -12} isn't a real answer, so x = 2.

Re: 10th grade math question--help me [oldandslow]
[ In reply to ]

oldandslow wrote:

The way tha tyou explained it, you have 8x12 => 96 sq. ft. of red roses. The total area will be twice that, 192 sq. ft. Which means that the new plot would be 12x16, with a two foot border around the original plot.^This is the way I read it.

You'll need to solve a quadratic equation. Let x = the width of the border. If the inner border is 8x12 it has an area of 96, but that also means the

Then you just solve (x + 12 + x)(x + 8 + x) - 96 = 96. In words, the area of the new rectangle created less the area of the already existing inside rectangle equals the area of the new border.

(2x+12)(2x+8)-96=96

(2x+12)(2x+8)=192

4x^2+40x+96=192

4x^2+40x-96=0

x^2+10x-24=0

(x+12)(x-2)=0

x= -12 or x = 2

So, x =2 and the width of the border is 2.

Thanks all!!!

I'm an idiot.

I'm an idiot.