The length of a rectangular poster is 9 more inches than three times its width. The area of the poster is 264 square inches . Solve for the dimensions (length and width) of the poster.

Let's denote the width of the rectangular poster as \( w \) inches. According to the problem, the length poster is 9 inches more than three times its width. Therefore, the length can be expressed as \( 3w + 9 \) inches.<br /><br />The area of the rectangle is given by the product of its length and width, which is 264 square inches. So, we can set up the equation:<br /><br />\[ w \times (3w + 9) = 264 \]<br /><br />Expanding the equation, we get:<br /><br />\[ 3w^2 + 9w = 264 \]<br /><br />To solve this quadratic equation, we can rearrange it to standard form:<br /><br />\[ 3w^2 + 9w - 264 = 0 \]<br /><br />Now, we can solve this quadratic equation using the quadratic formula:<br /><br />\[ w = \frac{-b \pm \sqrt{b^2 - 42a} \]<br /><br />In this case, \( a = 3 \), \( b = 9 \), and \( c = -264 \). Plugging these values into the quadratic formula, we get:<br /><br />\[ w = \frac{-9 \pm \sqrt{9^2 - 4 \times 3 \times (-264)}}{2 \times 3} \]<br /><br />\[ w = \frac{-9 \pm \sqrt{81 + 3168}}{6} \]<br /><br />\[ w = \frac{-9 \pm \sqrt{3249}}{6} \]<br /><br />\[ w = \frac{-9 \pm 57}{6} \]<br /><br />Now, we have two possible solutions for \( w \):<br /><br />\[ w_1 = \frac{-9 + 57}{6} = \frac{48}{6} = 8 \]<br /><br />\[ w_2 = \frac{-9 - 57}{6} = \frac{-66}{6} = -11 \]<br /><br />Since the width cannot be negative, we discard \( w_2 \) and take \( w = 8 \) inches as the width of the poster.<br /><br />Now, we can find the length of the poster using the expression \( 3w + 9 \):<br /><br />\[ \text{Length} = 3 \times 8 + 9 = 24 + 9 = 33 \text{ inches} \]<br /><br />Therefore, the dimensions of the rectangular poster are 8 inches in width and 33.