document.write( "Question 568702: Model with a quadratic equation, then solve.
\n" ); document.write( "Find the dimensions of a square picture that make the area of the picture equal to 75% of the total area enclosed by the frame. The total length of the side is 12inches, including the frame.
\n" ); document.write( "I thought y=axsquared for the equation
\n" ); document.write( "144=0.75x2
\n" ); document.write( "192=x2
\n" ); document.write( "x=13.9 But this is larger than the length of the picture frame???
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Algebra.Com's Answer #367116 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!
You've got it backwards.
\n" ); document.write( "If the length of the side of the picture frame is $\"12\"$ inches, the total area of frame plus picture is
\n" ); document.write( " $\"12%5E2\"$ square inches = $\"144\"$ square inches.
\n" ); document.write( "If $\"x\"$ inches is the length of the side of the square picture, the area of the square picture is $\"x%5E2\"$ square inches. Since that is 75% of the $\"144\"$ square inches of total area,
\n" ); document.write( " $\"x%5E2=0.75%2A144\"$ --> $\"x%5E2=108\"$
\n" ); document.write( "Solving:
\n" ); document.write( " $\"x=sqrt%28108%29\"$ --> $\"x=sqrt%284%2A27%29\"$ --> $\"x=sqrt%282%5E2%2A3%5E3%29\"$ --> $\"x=sqrt%282%5E2%29%2Asqrt%283%5E3%29\"$ --> $\"x=sqrt%282%5E2%29%2Asqrt%283%5E2%2A3%29\"$ --> $\"x=2%2Asqrt%283%5E2%29%2Asqrt%283%29\"$ --> $\"x=2%2A3%2Asqrt%283%29\"$ --> $\"x=6sqrt%283%29\"$ \n" ); document.write( "

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