Chapter 8.7, Problem 43PPE

To determine

To find: The expression representing the difference of areas of two rugs.

Answer to Problem 43PPE

The expression for the difference of areas of two rugs is $A_1-A_2=33x-80x+7$ .

Explanation of Solution

Given information:

A square rug has an area $49x^2-56x+16$ and another rug has an area $16x^2+24x+9$ . The difference of areas of rugs is to be find in two ways.

Formula used:

We solve the inequality $v\ge 8\text{cm/sec}$ .

Calculations:

Method 1:

The area $A_1$ (say) of first rug is $A_1=49x^2-56x+16$.........(1)

The area $A_2$ (say) of first rug is $A_2=16x^2+24x+9$.........(2)

The expression for the difference of areas of two rugs is

  $A_1-A_2=\left(49x^2-56x+16\right)-\left(16x^2+24x+9\right)$

  $\Rightarrow A_1-A_2=49x^2-16x^2-56x-24x+16-9$

  $\Rightarrow A_1-A_2=33x^2-80x+7$.........(3)

Method 2:

The area $A_1$ (say) of first rug is $A_1=49x^2-56x+16$ that may be written as

  $A_1=49x^2-56x+16={\left(7x\right)}^2-2\cdot 7x\cdot 4+{\left(4\right)}^2$

  $A_1={\left(7x-4\right)}^2$

The area $A_2$ (say) of first rug is $A_2=16x^2+24x+9$ that may be written as

  $A_2=16x^2+24x+9={\left(4x\right)}^2+2\cdot 4x\cdot 3+{\left(3\right)}^2$

  $A_2={\left(4x+3\right)}^2$ .

The expression for the difference of areas of two rugs is

  $A_1-A_2={\left(7x-4\right)}^2-{\left(4x+3\right)}^2$

Difference of two squares

  $A_1-A_2=\left(7x-4-4x-3\right)\left(7x-4+4x+3\right)$$\because a^2-b^2=\left(a-b\right)\left(a+b\right)$

  $A_1-A_2=\left(3x-7\right)\left(11x-1\right)$

  $A_1-A_2=33x-3x-77x+7$

  $A_1-A_2=33x-80x+7$.........(4)

The results (3) and (4) are same.

Hence, the expression for the difference of areas of two rugs is $A_1-A_2=33x-80x+7$ .

Conclusion:

The expression for the difference of areas of two rugs is $A_1-A_2=33x-80x+7$ .