Given:
The area of a rectangle is 99 square yd.
Length of the rectangle = 7 yd more than twice the width.
To find:
The dimensions of the rectangle.
Solution:
Let x be the width of the rectangle. Then, length of the rectangle is:
[tex]Length=2x+7[/tex]
Area of a rectangle is:
[tex]A=length\times width[/tex]
[tex]A=(2x+7)\times x[/tex]
[tex]A=2x^2+7x[/tex]
The area of a rectangle is 99 square yd.
[tex]2x^2+7x=99[/tex]
[tex]2x^2+7x-99=0[/tex]
Splitting the middle term, we get
[tex]2x^2+18x-11x-99=0[/tex]
[tex]2x(x+9)-11(x-9)=0[/tex]
[tex](x+9)(2x-11)=0[/tex]
Using zero product property, we get
[tex](x+9)=0[/tex] and [tex](2x-11)=0[/tex]
[tex]x=-9[/tex] and [tex]x=\dfrac{11}{2}[/tex]
[tex]x=-9[/tex] and [tex]x=5.5[/tex]
Width of the rectangle cannot be negative. So, [tex]x=5.5[/tex] yd.
Now, the length of the rectangle is:
[tex]Length=2x+7[/tex]
[tex]Length=2(5.5)+7[/tex]
[tex]Length=11+7[/tex]
[tex]Length=18[/tex]
Therefore, the length of the rectangle is 18 yd and the width of the rectangle is 5.5 yd.
