Answer:
The width is:
[tex]-2+\sqrt{26}\text{ meters}\text{ }(\text{or approximately 3.0990 meters})[/tex]
And the length is:
[tex]2+\sqrt{26}\text{ meters}\text{ } (\text{or approximately 7.0990 meters})[/tex]
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 of a rectangle is four meters longer than the width. Thus:
[tex]\ell = w + 4[/tex]
And we also know that the area of the rectangle is 22 square meteres.
Substitute:
[tex](22)=w(w+4)[/tex]
Distribute and isolate the equation:
[tex]w^2+4w-22=0[/tex]
The equation isn't factorable, so we can instead use the quadratic formula:
[tex]\displaystyle w = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In this case, a = 1, b = 4, and c = -22. Substitute:
[tex]\displaystyle w = \frac{-(4)\pm\sqrt{(4)^2-4(1)(-22)}}{2(1)}[/tex]
Evaluate:
[tex]\displaystyle\begin{aligned} w &= \frac{-4\pm\sqrt{104}}{2}\\ \\ &=\frac{-4\pm\sqrt{4\cdot 26}}{2} \\ \\ &=\frac{-4\pm2\sqrt{26}}{2} \\ \\ & = -2\pm \sqrt{26} \end{aligned}[/tex]
Thus, our two solutions are:
[tex]w_1=-2+\sqrt{26}\approx 3.0990\text{ or } w_2=-2-\sqrt{26}\approx-7.0990[/tex]
Since the width cannot be negative, we can ignore the second solution.
Since the length is four meters longer than the width:
[tex]\ell = (-2+\sqrt{26})+4=2+\sqrt{26}\text{ meters}[/tex]
Thus, the dimensions of the rectangle are:
[tex]\displaystyle (2+\sqrt{26}) \text{ meters by } (-2+\sqrt{26})\text{ meters}[/tex]
Or, approximately 3.0990 by 7.0990.
