Answer:
Length, l = 11 ft.
Width, w = 9 ft.
Step-by-step explanation:
From the given data, the area of the rectangle = 99 ft².
Area of the rectangle = Length, l X Width, w
Here, Length, l = 7 more than twice the width
⇒ Length, l = 7 + 2w
Therefore, Area, A = 99 = (7 + 2w)w
⇒ 99 = 7w + 2w²
⇒ 2w² + 7w - 99 = 0
Solve the Quadratic equation using the formula: x = [tex]$ \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $[/tex] for the quadratic equation ax² + bx + c = 0.
Therefore, w = [tex]$ \frac{-7 \pm \sqrt{7^2 -4(2)(-99)}}{2(2)} $[/tex]
[tex]$ = \frac{-7 \pm \sqrt{49 + 8(99)}}{4} $[/tex]
[tex]$ = \frac{-7 \pm + \sqrt{841}}{4} $[/tex]
Since, [tex]$ \sqrt{841} = 29 $[/tex] we get:
[tex]$ w = \frac{-7 \pm 29}{4} $[/tex]
This gives two values of 'w', viz., w = [tex]$ \frac{-7 - 29}{4} $[/tex], [tex]$ \frac{-7 + 29}{4} $[/tex]
[tex]$ \implies w = \frac{22}{4}, \frac{-36}{4} $[/tex]
⇒ w = [tex]$ \frac{22}{4} $[/tex], -9.
We take the integer values.
If w = -9, then l = 2(-9) + 7
⇒ l = - 18 + 7 = - 11
Therefore, the length, l of the rectangle = - 11 ft.
and the width, w of the rectangle = - 9 ft.
Hence, the answer.
