Question

A rectangle has a length that is nine feet less than four times it's width it's area is 90 square feet algebraiclly determine the length of its width and length

Answer 1

Answer:

Length of the rectangle is 15 feet and width of the rectangle is 9 feet.

Step-by-step explanation:

Given:

Area of the rectangle = 90 square feet

We need to find the length and width of the rectangle.

Solution:

Let the width of the rectangle be denoted by 'w'.

Now given:

length is nine feet less than four times it's width.

so we can say that;

length of the rectangle = $4w-9$

Now we know that;

Area of the rectangle is given by length times width.

framing in equation form we get;

$(4w-9)w=90\\\\4w^2-9w =90\\\\4w^2-9w-90=0$

Now we will factorize so as to find the roots.

$4w^2-24w+15w -90=0\\\\4w(w-6)+15(w-6)=0\\\\(4w+15)(w-6)=0$

Now substituting separately we will find 2 value of w.

$4w+15 =0 \ \ \ \ Or \ \ \ \ w-6 =0\\\\4w=-15 \ \ \ \ \ Or \ \ \ \ \ w= 6\\\\w =\frac{-15}{4} \ \ \ \ \ Or \ \ \ \ \ w= 6$

Now we get 2 value of w one positive and one negative and we know that width of the rectangle can't be negative hence we will discard negative value  and consider positive value.

width of the rectangle = 6 ft

Length of the rectangle = $4x-9 = 4\times6 -9 = 24 -9 =15\ ft$

Hence Length of the rectangle is 15 feet and width of the rectangle is 9 feet.