Difference of Squares Example Problems

Solution:

Notice that ${\mathrm{a}}^{100}-{\mathrm{b}}^{100}$ is in the form of a difference of squares because a¹⁰⁰ is the square of a⁵⁰ and b¹⁰⁰ is the square of b⁵⁰.

Thus, ${\mathrm{a}}^{100}-{\mathrm{b}}^{100}={\left({\mathrm{a}}^{50}\right)}^2-{\left({\mathrm{b}}^{50}\right)}^2=\left({\mathrm{a}}^{50}-{\mathrm{b}}^{50}\right)\left({\mathrm{a}}^{50}+{\mathrm{b}}^{50}\right)$.

$\begin{array}{l}\Rightarrow \frac{{\mathrm{a}}^{100}-{\mathrm{b}}^{100}}{{\mathrm{a}}^{50}-{\mathrm{b}}^{50}}=\frac{{\left({\mathrm{a}}^{50}\right)}^2-{\left({\mathrm{b}}^{50}\right)}^2}{{\mathrm{a}}^{50}-{\mathrm{b}}^{50}}\\ =\frac{\left(\bcancel{{\mathrm{a}}^{50}-{\mathrm{b}}^{50}}\right)\left({\mathrm{a}}^{50}+{\mathrm{b}}^{50}\right)}{\bcancel{{\mathrm{a}}^{50}-{\mathrm{b}}^{50}}}\\ ={\mathrm{a}}^{50}+{\mathrm{b}}^{50}\end{array}$

Solution:

It would take a significant amount of time to square the values in the numerator and then subtract them. An easier approach is to notice that the numerator is in the form of a difference of squares:

${\mathrm{x}}^2-{\mathrm{y}}^2=\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}-\mathrm{y}\right)$. Here, x = 555, and y = 55.

$\begin{array}{l}\Rightarrow \frac{{\left(555\right)}^2-{\left(55\right)}^2}{500}\\ \Rightarrow \frac{\left(555+55\right)\left(555-55\right)}{500}\\ \Rightarrow \frac{\left(610\right)\left(\cancel{500}\right)}{\cancel{500}}\\ \Rightarrow 610\end{array}$

Solution:

The difference of the areas can be expressed as $\mathrm{\pi }{\mathrm{R}}^2-\mathrm{\pi }{\mathrm{r}}^2$. We can use the difference of squares to help us easily subtract these quantities. 

$\begin{array}{l}\Rightarrow \mathrm{\pi }{\mathrm{R}}^2-\mathrm{\pi }{\mathrm{r}}^2\\ \Rightarrow \mathrm{\pi }\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)\\ \Rightarrow \mathrm{\pi }\left(\mathrm{R}-\mathrm{r}\right)\left(\mathrm{R}+\mathrm{r}\right)\\ \Rightarrow \mathrm{\pi }\left(555-55\right)\left(555+55\right)\\ \Rightarrow \mathrm{\pi }\left(500\right)\left(610\right)\\ \Rightarrow 305,000\mathrm{\pi }\end{array}$

$\mathrm{305,000}\text{π}$ is the difference between the areas of the two circles in square centimeters. However, we want to know the difference between the areas in square meters. Since there are 100 centimeters in 1 meter, there are 100² = 10,000 square centimeters in 1 square meter.  Therefore, the difference between the areas of the two circles, in meters, is: $30.5\text{π}$.

Solution:

To mark the item’s price up x percent, we must remember that we are increasing the item’s original price of d dollars by x percent. The percent increase function is expressed as $\left(1+\frac{\mathrm{x}}{100}\right)$. Notice that without adding 100% to the x percent, we would be reducing the item’s price by (100 – x) percent. Similarly, when reducing the item’s resulting price by x percent, we must multiply this new starting price by $\left(1-\frac{\mathrm{x}}{100}\right)$. The item’s ending price is therefore

$\begin{array}{l}\Rightarrow \mathrm{d}\left(1+\frac{\mathrm{x}}{100}\right)\left(1-\frac{\mathrm{x}}{100}\right)\\ \Rightarrow \mathrm{d}\left(\frac{100+\mathrm{x}}{100}\right)\left(\frac{100-\mathrm{x}}{100}\right)\\ \Rightarrow \mathrm{d}\left(\frac{{100}^2-{\mathrm{x}}^2}{\mathrm{10,000}}\right)\\ \Rightarrow \mathrm{d}\left(\frac{\mathrm{10,000}-{\mathrm{x}}^2}{\mathrm{10,000}}\right)\end{array}$

As we can see in the calculation above, there is a difference of squares of $\left(100+\mathrm{x}\right)\left(100-\mathrm{x}\right)$ in the numerators of the fractions. We can therefore avoid the standard method of simplifying a quadratic equation given the property $\left(100+\mathrm{x}\right)\left(100-\mathrm{x}\right)=\left({100}^2-{\mathrm{x}}^2\right)$.

Solution:

Our job is to determine which terms represent the largest and smallest values in the set. Since three of the six terms involve x and the other three involve y, we need to find the largest and the smallest terms that involve x and the largest and the smallest terms that involve y.

We know that x is an integer greater than one, so it must be positive. Out of all the values for terms involving x, x² must be the greatest because the square of an integer greater than one will always be greater than the integer, and $\frac{1}{4}\mathrm{x}$ must be the smallest since $\frac{1}{4}$ of a positive number is less than the number. For example, if x = 2 (the smallest integer greater than 1), then x² = 4, and $\frac{1}{4}\mathrm{x}=0.5$. Notice that 4 and 0.5 are the smallest values of x² and $\frac{1}{4}\mathrm{x}$, respectively.

Next, we know that y is a positive decimal less than or equal to 0.7. Out of all the values for terms involving y, 5y must be the greatest because 5 times a positive number is greater than the number, and y² must be the smallest since the square of a positive decimal less than one will always be less than the decimal. For example, if y = 0.7 (the largest decimal it can be), then 5y = 3.5 and y² = 0.49. Notice that 3.5 and 0.49 are the largest values of 5y and y², respectively.

As we can see, for the term with the largest value, we should either choose x² or 5y since they are the greatest of the terms involving x or y, respectively. However, we should choose x² since the smallest possible value of x² is 4, which is greater than 3.5, the largest possible value of 5y. Similarly, for the term with the smallest value, we should either choose $\frac{1}{4}\mathrm{x}$ or y² since they are the smallest of the terms involving x or y, respectively. However, we should choose y² since the largest possible value of y² is 0.49, which is less than 0.5, the smallest possible value of $\frac{1}{4}\mathrm{x}$.

Consequently, the range equals x² – y², which in this case is represented as the difference of squares in factored form as (x + y)(x – y).