Answer:
[tex]4+2\sqrt{31}\text{ by } -4+2\sqrt{31}[/tex]
Or about 15.136 centimeters by 7.136 centimeters.
Step-by-step explanation:
Recall that the area of a rectangle is given by:
[tex]\displaystyle A = w\ell[/tex]
Where w is the width and l is the length.
We are given that the length is 8 centimeters longer than the width. In other words:
[tex]\ell = w+8[/tex]
And we are also given that the total area is 108 square centimeters.
Thus, substitute:
[tex](108)=w(w+8)[/tex]
Solve for w. Distribute:
[tex]w^2+8w=108[/tex]
Subtract 108 from both sides:
[tex]w^2+8w-108=0[/tex]
Since the equation is not factorable, we can use the quadratic formula:
[tex]\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In this case, a = 1, b = 8, and c = -108. Substitute and evaluate:
[tex]\displaystyle \begin{aligned} w&= \frac{-(8)\pm\sqrt{(8)^2-4(1)(-108)}}{2(1)} \\ \\ &=\frac{-8\pm\sqrt{496}}{2}\\ \\ &=\frac{-8\pm4\sqrt{31}}{2} \\ \\ &=-4 \pm 2\sqrt{31} \end{aligned}[/tex]
So, our two solutions are:
[tex]w=-4+2\sqrt{31} \approx 7.136 \text{ or } w=-4-2\sqrt{31}\approx -15.136[/tex]
Since width cannot be negative, we can eliminate the second solution.
And since the length is eight centimeters longer than the width, the length is:
[tex]\ell =(-4+2\sqrt{31})+8=4+2\sqrt{31}\approx 15.136[/tex]
So, the dimensions of the rectangle are about 15.136 cm by 7.136 cm.
