SOLUTION: A square poster had 9 in. added to its width and 2 in. subtracted from its height. The poster then had an area of 102in.to the 2nd power.How long was a side of the original squa
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-> SOLUTION: A square poster had 9 in. added to its width and 2 in. subtracted from its height. The poster then had an area of 102in.to the 2nd power.How long was a side of the original squa
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Question 32473: A square poster had 9 in. added to its width and 2 in. subtracted from its height. The poster then had an area of 102in.to the 2nd power.How long was a side of the original square poster? Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website! (X+9)(X-2)=102 OR X~2+7X-18=102 OR X~2+7X-120=0 OR (X+15)(X-8)=0 OR X=8
PROOF (8+9)(8-2)=102 OR 17*6=102 OR 102=102
ORIGINAL SIDES WERE 8IN.
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