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The area of a rectangle is 42 ft squared, and the length of the rectangle is 5 ft more than twice the width. Find the dimensions of the rectangle. length and width.​

Respuesta :

TheAnimeGirl

Answer:

Length = 12 ft

Width = [tex] \frac{7}{2} ft[/tex]

Step-by-step explanation:

Given,

Area of rectangle = [tex]42 \: {ft}^{2} [/tex]

Width = X

Length = 2x + 5

Now,

[tex]x(2x + 5) = 42[/tex]

[tex]2 {x}^{2} + 5x = 42[/tex]

[tex]2 {x}^{2} + 5x - 42 = 0[/tex]

[tex]2 {x}^{2} + 12x - 7x - 42 = 0[/tex]

[tex]2x(x + 6) - 7(x + 6) = 0[/tex]

[tex](2x - 7)(x + 6) = 0[/tex]

Either

[tex]2x - 7 = 0[/tex]

[tex]2x = 0 + 7[/tex]

[tex]2x = 7[/tex]

[tex]x = \frac{7}{2} [/tex]

Or,

[tex]x + 6 = 0[/tex]

[tex]x = 0 - 6[/tex]

[tex]x = - 6[/tex]

Negative value can't be taken.

So, width = [tex] \frac{7}{2} ft[/tex]

Again,

Finding the value of length,

Length = [tex]2x + 5[/tex]

[tex]2 \times \frac{7}{2} + 5[/tex]

[tex]7 + 5[/tex]

[tex]12[/tex]

Length = 12 ft

valerieulbrich

Answer:

length = 12 ft, width = 3.5 ft

Step-by-step explanation:

w = width

l = length = 2w + 5

A = wl = w(2w + 5) = 42

2w² + 5w - 42 = 0

(w + 6)(2w - 7) = 0

w + 6 = 0, w = -6 (dimension cannot be negative)

2w - 7 = 0, w = 3.5

l = 2(3.5) + 5 = 12

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