   Chapter A.5, Problem 33E Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Solutions

Chapter
Section Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

The length of a rectangle is 3 more than its width. If the area of the rectangle is 40, its dimension x     a n d     x + 3 can be found by solving the equation x ( x + 3 ) = 40 . Find these dimension.

To determine

To find:

To find the dimension of length and width by using the equation x(x+3)=40

Explanation

Consider the following condition,

“The length of a rectangle is 3 more than its width. If the area of the rectangle is 40, its dimension xandx+3 can be found by solving the equation x(x+3)=40”.

Let x be length of the rectangle and x+3 be width of the rectangle.

Since length of a rectangle is 3 more than its width.

That is,

l=xw=x+3

Area of the rectangle is 40.

Given that x(x+3)=40

Then simplify the above expression by multiplying the LHS to get the following,

x(x+3)=40x2+3x=40

Then subtract 40 on both sides to get the following,

x2+3x40=4040x2+3x40=0

The general form of the quadratic equation is given below,

ax2+bx+c=0

Where a is the number multiplied by x2, b is the number multiplied by x, and c is the constant term.

The solution for ax2+bx+c=0 when a0 is given below,

x=b±b24ac2a...(1)

Compare the given equation with general equation to get the following,

ax2+bx+c=0x2+3x40=0

Therefore,

a=1,b=3andc=40. Since a0.

Substitute these values in equation (1) to get the following,

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