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3. 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?​


Sagot :

Answer:

42944%÷×+59864314476=×

Step-by-step explanation:

584×÷2680=+

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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