The length of a rectangle is 1 foot less than 4 times its width. How do you find the dimensions of this rectangle if the area is 60 square feet?

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The length of a rectangle is 1 foot less than 4 times its width. How do you find the dimensions of this rectangle if the area is 60 square feet?

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Answer 1 · 2016-02-10T13:03:37

Dimensions of rectangle are $4$ $4$ and $15$ $15$ .

Explanation:

Assume that the width in feet is $x$ $x$ . Hence, length of rectangle, which is one less than four times, will be $4 x - 1$ $4x-1$ .

As area of a rectangle (which is $60$ $60$ ) is length multiplied by width, it is given by $x \cdot (4 x - 1)$ $x*(4x-1)$ i.e. $4 x^{2} - x$ $4x^2-x$ .

Hence $4 x^{2} - x = 60$ $4x^2-x=60$ or

$4 x^{2} - x - 60 = 0$ $4x^2-x-60=0$ .

As this is a quadratic equation of type $a \cdot x^{2} + b \cdot x + c$ $a*x^2+b*x+c$ , to factorize this as $a \cdot c$ $a*c$ is negative (it is $- 240$ $-240$ ), we should factorize $240$ $240$ in two factors whose difference is $1$ $1$ note that $b = - 1$ $b=-1$ . Hit and trial shows these are 16 and 15.

Hence equation can be written as

$4 x^{2} - 16 x + 15 x - 60 = 0$ $4x^2-16x+15x-60=0$ or $4 x (x - 4) + 15 (x - 4) = 0$ $4x(x-4)+15(x-4)=0$

or $(4 x + 5) \cdot (x - 4) = 0$ $(4x+5)*(x-4)=0$ i.e.

solution is $x = - \frac{5}{4}$ $x=-5/4$ or $x = 4$ $x=4$ .

As width cannot be $- \frac{5}{4}$ $-5/4$ , it is $4$ $4$ and length is $(4 \cdot 4 - 1)$ $(4*4-1)$ or $15$ $15$ .

Hence dimensions of rectangle are $4$ $4$ and $15$ $15$ .

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The length of a rectangle is 1 foot less than 4 times its width. How do you find the dimensions of this rectangle if the area is 60 square feet?

1 Answer

Explanation:

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