document.write( "Question 1188666: Mr.Ulep has a rectangular field with a length twice its width.If 4 feet are taken from the length and 3 feet are taken from the width, the area of the field is 180 square feet.Find the area of the original field \n" ); document.write( "

Algebra.Com's Answer #819797 by Shin123(626) $\"\"$ $\"About$

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Since the length is twice the width, let's say the length is $\"2x\"$ and the width is $\"x\"$ . So we have the equation $\"%282x-4%29%28x-3%29=180\"$ . Expanding, we get $\"2x%5E2-10x-168=0\"$ .
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Solved by pluggable solver: Completing the Square for Quadratics

To complete the square for the quadratic $\"2%2Ax%5E2-10%2Ax-168=0\"$ , we must first find a square which when expanded, has 2x² and -10x in it.
\n" ); document.write( "Factoring 2 from the left side gives $\"2%28x%5E2-5%2Ax-84%29=0\"$ . $\"%28x-2.5%29%5E2\"$ is the square we are looking for. So we get $\"2%28%28x-2.5%29%5E2-90.25%29=0\"$ .\n" ); document.write( "Taking the -90.25 out of the 2, we get $\"highlight%282%28x-2.5%29%5E2-180.5%29\"$ . Adding 180.5 to both sides, we get $\"2%28%28x-2.5%29%5E2%29=180.5\"$ . Dividing both sides by 2 gives $\"%28x-2.5%29%5E2=90.25\"$ . Taking the square root of both sides gives $\"system%28x-2.5=sqrt%2890.25%29=9.5%2Cx-2.5=-sqrt%2890.25%29=-9.5%29\"$ . $\"system%28x=12%2Cx=-7%29\"$ So the solutions are x=12 and x=-7.

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\n" ); document.write( "Since the side lengths can't be negative, we discard $\"x=-7\"$ , so x must be 12.
\n" ); document.write( "Therefore, the area is $\"%282%2A12%29%2A12=highlight%28highlight_green%28highlight%28288%29%29%29\"$ ft². \n" ); document.write( "

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