Problem:
The length of the rectangular bulletin board is 2 meters more than its width. If its area is less than 7m², what will be the possible width?
Solution:
L = 2 + W
A = 7 m²
A = LW
7 = (2 + W)(W)
W² + 2W - 7 = 0
[tex]\[\begin{array}{l}W1,W2 = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\\\W1,W2 = \frac{{ - 2 \pm \sqrt {{2^2} - 4(1)( - 7)} }}{{2(1)}}\\\\\sqrt {{b^2} - 4ac} \\\\\sqrt {{2^2} - 4(1)( - 7)} \\\\\sqrt {4 + 28} \\\\\sqrt {32} \\\\W1,W2 = \frac{{ - 2 \pm \sqrt {32} }}{{(2)}}\\\\W1 = \frac{{ - 2 + \sqrt {32} }}{{(2)}}\\\\W1 = {\rm{1}}{\rm{.828427124746}}m\\\\W2 = \frac{{ - 2 - \sqrt {32} }}{{(2)}}\\\\W2 = - {\rm{3}}{\rm{.8284271247m}}\end{array}\][/tex]
use the positive value of W = 1.828427124746m
L = 2 + W
L = 2 + 1.828427124746
L = 3.828427124746m
W = 1.828427124746m
Checking:
A = LW
7 = (3.828427124746)(1.828427124746)
7 = 7 ; ok
Answer:
W = 1.828427124746m
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