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Someone help ASAP!!!!!!

Someone help ASAP!!!!!!-example-1

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Answer:

Length, l = 11 ft.

Width, w = 9 ft.

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 =
$ (-b \pm √(b^2 - 4ac))/(2a) $ for the quadratic equation ax² + bx + c = 0.

Therefore, w =
$ (-7 \pm √(7^2 -4(2)(-99)))/(2(2)) $


$ = (-7 \pm √(49 + 8(99)))/(4) $


$ = (-7 \pm + √(841))/(4) $

Since,
$ √(841) = 29 $ we get:


$ w = (-7 \pm 29)/(4) $

This gives two values of 'w', viz., w =
$ (-7 - 29)/(4) $,
$ (-7 + 29)/(4) $


$ \implies w = (22)/(4), (-36)/(4) $

⇒ w =
$ (22)/(4) $, -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.

User Bisjob
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