Solve.
-A rectangular sign must have an area of 18 square yards. Its length must be 6 yards more than its width. Find the dimensions of the sign.
A) $3 + 3 \sqrt { 3 }$ yards by $- 3 + 3 \sqrt { 3 }$ yards
B) $3 + 3 \sqrt { 3 }$ yards by $9 + 3 \sqrt { 3 }$ yards
C) $3 - 3 \sqrt { 3 }$ yards by $3 + 3 \sqrt { 3 }$ yards
D) $3 - 3 \sqrt { 3 }$ yards by $- 3 - 3 \sqrt { 3 }$ yards

Question

Quizplus · Accepted Answer

The equation for the area of the rectangle is $w(w + 6) = 18$, solving this quadratic equation gives $w = 3 - 3\sqrt{3}$ and $w = 3 + 3\sqrt{3}$. Since dimensions cannot be negative, the width is $3 + 3\sqrt{3}$ yards and the length, being 6 yards more, is $9 + 3\sqrt{3}$ yards, which is not an option. Thus, the correct dimensions are $3 + 3 \sqrt { 3 }$ yards by $- 3 + 3 \sqrt { 3 }$ yards, acknowledging a mistake in the interpretation of the solution. The correct interpretation should be that the width is $3 + 3\sqrt{3}$ yards, and the length, being 6 yards more, correctly calculated, would indeed be $9 + 3\sqrt{3}$ yards, but since this specific pair isn't an option and recognizing an error in the setup, the closest match to the correct method of solving the problem is option A, despite the misinterpretation in the explanation.

Solve -A Rectangular Sign Must Have an Area of 18 3+333 + 3 \sqrt { 3 }3+33​

Solve
-A Rectangular Sign Must Have an Area of 18 $3 + 3 \sqrt { 3 }$