Question 155157: Can someone help with the following word problem, I tried the quadratic equation but I am not getting any of the answers.
Thank you.
Hazel has a screen door whose height is 4 feet more than its width. She wishes to stabilize the door by attaching a steel cable diagonally. If the cable measures (sqrt 19)/2 ft, what are the dimensions of the door? (1 point)
A) 2 1/4 ft by 6 1/4 ft
B) 2 1/2 ft by 6 1/2 ft
C) 3 ft by 7 ft
D) 3 1/2 ft by 7 1/2 ft
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Hazel has a screen door whose height is 4 feet more than its width. She wishes to stabilize the door by attaching a steel cable diagonally. If the cable measures (sqrt 19)/2 ft, what are the dimensions of the door? (1 point)
A) 2 1/4 ft by 6 1/4 ft
B) 2 1/2 ft by 6 1/2 ft
C) 3 ft by 7 ft
D) 3 1/2 ft by 7 1/2 ft
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Draw the picture.
hypotenuse = [sqrt(19)]/2 ft.
Width : x ft.
height: x+4 ft.
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EQUATION:
x^2 + (x+4)^2 = [sqrt(19)/2]
x^2 + x^2 + 8x + 16 = 19/4
2x^2 + 8x + 11.25 = 0
x = [-8 +- sqrt(64 - 4*2*11.25)]/4
x = [-8 +- sqrt(-26)]4
Note: This value is not a Real Number.
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Something is wrong in the statement of your problem.
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Cheers,
Stan H.
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