Jessica has a picture in a frame with a total area of 288 in2. The dimension of the picture without the frame is 12 in through 14 in. What is the larger dimension, in inches, of the frame?
To solve this problem, ?nd out the width of the frame ?rst. Let x = the width of the frame. Since the width of the picture just is 12, the width of the frame and the picture is x + x + 12 or 2x + 12. Because the length of the picture only is 14, the length of the frame and the picture together is x + x + 14 or 2x + 14. The total field for the frame and the picture together is 288 square inches. Area of a rectangle is length times width so multiply the expressions jointly and set them equal to the total area of 288 square inches: (2x + 12)(2x + 14) = 288. Multiply the binomials by using the distributive property: 4x2 + 28x + 24x + 168 = 288. Combine like terms: 4x2 + 52x + 168 = 288. Subtract 288 from both sides: 4x2 + 52x + 168 - 288 = 288 - 288; simplify: 4x2 + 52x - 120 = 0. Factor the trinomial completely: 4(x2 + 13x - 30) = 0; 4(x - 2)(x + 15) = 0. Set each factor equal to zero and solve: 4 ≠ 0 or x - 2 = 0 or x + 15 = 0; x = 2 or x = -15. Reject the negative solution since you will not have a negative width. The width is 2 feet. Thus, the larger dimension of the frame is 2(2) + 14 = 4 + 14 = 18 inches.