# Quant: Difference of Two Squares

## Prime numbers

You may remember this formula, one of the sleekest factoring tricks in all of algebra:

\bold{y^2 - x^2 = (y + x)(y - x)}

This formula, called “the difference of two squares” formula, is a favorite of standardized test writers. A simple enough pattern: see if you can detect where it shows up in the following challenging problems.

**Q1**. \dfrac{1}{2 - \sqrt{3}}=

(A) 2 - \sqrt{3}

(B) 2 + \sqrt{3}

(C) \dfrac{1}{7}

(D) \dfrac{1}{2} - \dfrac{1}{\sqrt{3}}

(E) \dfrac{1}{2} + \dfrac{1}{\sqrt{3}}

**Q2**. What is the sum of a and b?

Statement 1: a = 4

Statement 2: \dfrac{b^2 - a^2}{b - a} = 7

(A) Statement 1 alone is sufficient but statement 2 alone is not sufficient to answer the question asked.

(B) Statement 2 alone is sufficient but statement 1 alone is not sufficient to answer the question asked.

(C) Both statements 1 and 2 together are sufficient to answer the question but neither statement is sufficient alone.

(D) Each statement alone is sufficient to answer the question.

(E) Statements 1 and 2 are not sufficient to answer the question asked and additional data is needed to answer the statements.

**Q3**. In the diagram above, \angle{A} = \angle{ABC}, \angle{CBD} = \angle{BDC}, and \angle{CBE} = 90\degree. If AE = 16 and DE = 4, what is the length of BE?

(A) 7

(B) 8

(C) 9

(D) 10

(E) 11

**Q4**. If x^2 - y^2 = 12 and x - y = 4, then x =

(A) 1.5

(B) 2.5

(C) 3.5

(D) 4.5

(E) 5.5

## Practice Problem Solutions

**Q1**. This involves a relatively sophisticated trick known as “multiplying by the conjugate.” When we have an expression of the form a + \sqrt{b}, the “conjugate” of this is a - \sqrt{b}. When we multiply a radical expression by its conjugate, we employ the difference of two squares. For example:

\dfrac{1}{2 - \sqrt{3}} = \dfrac{1}{2 - \sqrt{3}} \times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}} = \dfrac{2 + \sqrt{3}}{2^2 - (\sqrt{3})^2} = \dfrac{2 + \sqrt{3}}{4 - 3} = \dfrac{2 + \sqrt{3}}{1} = 2 + \sqrt{3}

This is answer **B**. BTW, the trick of multiplying by the conjugate is at the very outside edge of what you might be expected to do the hardest QUANT math problems.

**Answer = B**

**Q2**. The prompt of this DS problem is straightforward.

Statement #1 tells us a = 4, but we have no idea of b, so, by itself, this is not sufficient for finding the sum.

Statement #2 gives us a value for an algebraic expression that lends itself nicely to simplification.

7 = \dfrac{b^2 - a^2}{b - a} = \dfrac{(b-a)(b+a)}{b-a} = a + b

Thus, the sum is 7. Statement #2, by itself is sufficient.

**Answer = B**

**Q3**. This is a tricky one. Remember, it’s a diagram drawn to scale, so if all else fails, you can estimate (see this post). But, let’s solve this with math. The fact that \angle{A} = \angle{ABC} tells us \triangle{ABC} is isosceles, with AC = BC. The fact that \angle{CBD} = \angle{BDC} tells us \triangle{BCD} is **isosceles**, with BC = CD. The fact that \angle{CBE} = 90\degree means that (BC)^2 + (BE)^2 = (CE)^2. This means

(BE)^2 = (CE)^2 - (BC)^2

= (CE + BC)(CE - BC)

= (CE + AC)(CE - CD) (substitutions from the two isosceles triangles)

= (AE)(DE)

= (16)(4) = 64

so

BE = 8

**Answer = B**

**Q4**.
**Answer = C**