The first of 5 is: A rectangular garden is 20 ft longer than it is wide. Its area is 8000 ft{}^2. What are its dimensions?

Next #2:
The area of a rectangle is 16, and its diagonal is \sqrt{68}. Find its dimensions and perimeter.

(x^2)+(y^2)=\sqrt{68}
(x^2)+(y^2)=8.246
xy=16
y=16/x
(x^2)+(16/x)^2=8.246
(x^4)+(96/x)=8.246

#3:A box with a square base and no top is to be made from a square piece of carboard by cutting 4 in. squares from each corner and folding up the sides. The box is to hold 784 in{}^3. How big a piece of cardboard is needed?

not sure how to approach this

#4

x^2+9x-1=0
(x-9)x=1
x^2=9x+1
(x-9/2)^2-85/4=0

Solutions x=1/2(-9- sq root 85)
and x=1/2(-9 + sq root 85)

The first portion was marked correctly, however the second solution was marked incorrect. I'm not sure why, please help me figure out why solution #2 is wrong.

Lastly, #5

2x^2+17x+2=0
x^2+17/2 x+ 1=0
x^2+17/2 x= -1

I know to get my final answer I must complete the square but I am not sure how to do so. This is a far as i got before getting stuck.

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3. 👁 168
1. I've figure out number #1 so I no longer need assistance on that one. But please help me with the following 4 questions.

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2. 1. Width = W ft.
Length = (W+20) ft.

A = W*(W+20) = 8,000 Ft^2
W^2 + 20W - 8000 = 0
C=-8,000 = -80*100. Sum = -80+100=20=B.
(w+100)(w-80) = 0

w+100 = 0, W = -100
w-80 = 0, W = 80 Ft = Width.
w+20 = 80+20 = 100 Ft = Length.

2. A = L*W = 16, L = 16/W.

Eq2: L^2 + W^2 = 68
In Eq2, replace L with 16/w:
(16/W)^2 + W^2 = 68
256/W^2 + W^2 = 68
Multiply by W^2:
256 + W^4 = 68W^2
W^4 - 68W^2 + 256 = 0
It was found by trial and error that 2
and -2 satisfied the 4th degree Eq.
W = 2.
L = A/w = 16/2 = 8
P = 2L + 2W = 2*8 + 2*2 = 20

4. x^2 + 9x - 1 = 0
X = 0.10977, and -9.1098

5. 2x^2 + 17x + 2 = 0
X = -0.11932, and -8.38068.

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posted by Henry

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