# Difference of Squares Example Problems

Try the the following 5 sample questions, hand-picked by the Target Test Prep GRE experts. When solving, use the difference of squares formula to answer each question strategically.

Which of the following is equivalent to $\frac{{\mathrm{a}}^{100}\text{}-\text{}{\mathrm{b}}^{100}}{{\mathrm{a}}^{50}\text{E}-\text{}{\mathrm{b}}^{50}}$ for all values of a and b for which the expression is defined?

Notice that ${\mathrm{a}}^{100}-{\mathrm{b}}^{100}$ is in the form of a difference of squares because a^{100} is the square of a^{50} and b^{100} is the square of b^{50}.

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

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

What is the value of $\frac{{\left(555\right)}^{2}-{\left(55\right)}^{2}}{500}$?

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.

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

The radius, R, of a larger circle is 555 centimeters. The radius, r, of a smaller circle is 55 centimeters. Which of the following properly expresses, in square meters, the difference between the areas of the two circles? (Note: 1 meter = 100 centimeters)

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.

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

$305\text{\pi}$ 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 100^{2} = 10,000 square centimeters in 1 square meter. Therefore, the difference between the areas of the two circles, in meters, is: $30.5\text{\pi}$.

On Monday an item was marked up x percent. On Tuesday it was marked down x percent. If the original price of the item was d dollars, which of the following correctly expresses the final price of the item after the markup and markdown?

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

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

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 $\left(100+\mathrm{x}\right)\left(100-\mathrm{x}\right)=\left({100}^{2}-{\mathrm{x}}^{2}\right)$.

Set S = {x, $\frac{1}{4}\mathrm{x}$, y^{2}, 5y, x^{2}, y}

If in set S above, x is an integer greater than one and 0 < y $\le $ 0.7, which of the following represents the range of the set S?

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, x^{2} 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 x^{2} = 4, and $\frac{1}{4}\mathrm{x}=0.5$. Notice that 4 and 0.5 are the smallest values of x^{2} 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 y^{2} 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 y^{2} = 0.49. Notice that 3.5 and 0.49 are the largest values of 5y and y^{2}, respectively.

As we can see, for the term with the largest value, we should either choose x^{2} or 5y since they are the greatest of the terms involving x or y, respectively. However, we should choose x^{2} since the smallest possible value of x^{2} 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 y^{2} since they are the smallest of the terms involving x or y, respectively. However, we should choose y^{2} since the largest possible value of y^{2} is 0.49, which is less than 0.5, the smallest possible value of $\frac{1}{4}\mathrm{x}$.

Consequently, the range equals x^{2} – y^{2}, which in this case is represented as the difference of squares in factored form as (x + y)(x – y).