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Explain why the method of solving AND the best answer (selected by the asker) to this question are incorrect?
The diagonal of a rectangular field measures 120 ft and the length is 3 ft more than twice the width. Find the length and width of the field.
Answer selected as best by asker:
(d^2) = (L^2) + (w^2) = (120^2)
L = 2W + 3
Substituting
(L^2) + (w^2) = (120^2)
(2w + 3)^2 + w^2 = 120^2
4w^2 + 9 + 12W + w^2 = 14,400
5w^2 + 12w - 14301 = 0
Solving the quadratic equation
w = 52.29 and -54.69
Skipping the negative one
w = 52.29
L = [(2)*(52.29 + 3)] + 3 = 107.58
Checking
d^2 = [(107.58)^2] + [(52.29)^2] = 11,573 + 2,734
d = 119.6
close enough
Note: The correct answers are:
Width = 47.85 ft
Length = 110.05 ft
2 Answers
- billrussell42Lv 710 years agoFavorite Answer
I'm the one who provided this solution.
Let me check the "correct" answers
47.85² = 2289.622
110.05² = 12111.0025
sum = 14400.62
square root of that is 120.0026, correct
L = 2W+3 = 95.7+3 = 98.7 ≠ 110.05
NOT correct, it does not meet the "3 ft more than twice the width" requirement.
I can go through my numbers again and get a more accurate number
And I spotted an error. Quadratic should be
5w² + 12w – 14391 = 0
solutions are
w = 52.4622
L = 107.9245
checking
52.4622² = 2752.3
107.9245² = 11647.7
sqr of sum = 119.9999 OK
L = 2W+3 = 107.9244 OK
a lot closer than my original calculation, I subtracted incorrectly.
But I always say: check the math.
- ?Lv 710 years ago
I went through you work, but didn't actually put the number into the quadratic. It looks correct. The values you gave for what is supposed to be the correct answers aren't correct, because 2(47.8)+3 is not equal to 110.05.
