2-2 Problems 2) Juan wants to paint his bedroom wall. The length of the wall is 8ft. less than twice its width. If the area of the wall is 192ft^2 what are the dimensions of the wall? Level 3 - 3 Problems

To solve this problem, we need to find the dimensions of the wall based on the given information.<br /><br />Given information:<br />- The length of the wall is 8ft. less than twice its width.<br />- The area of the wall is 192ft².<br /><br />Let's define the variables:<br />- Let the width of the wall be x ft.<br />- Then, the length of the wall would be 2x - 8 ft.<br /><br />The area of the wall192ft², so we can write the equation:<br />Area = Length × Width<br />192 = (2x - 8) × x<br /><br />Now, we can solve this equation to find the value of x:<br />192 = 2x² - 8x<br />0 = 2x² - 8x - 192<br />0 = 2(x² - 4x - 96)<br />0 = 2(x - 12)(x + 8)<br /><br />The solutions to this equation are:<br />x = 12 or x = -8<br /><br />Since the width cannot be negative, the width of the wall is 12 ft.<br /><br />Now, we can find the length of the wall:<br />Length = 2x - 8<br />Length = 2(12) - 8<br />Length = 24 - 8<br />Length = 16 ft<br /><br />Therefore, the dimensionsWidth = 12 ft<br />Length = 16 ft