Worked Examples: GRE Algebra

Example 1: Perfect-Square Quadratic (Medium)

If $x^2 = 6x - 9$, what is the value of $x$?

Solution. Move everything to one side: $x^2 - 6x + 9 = 0$. Recognize the perfect square: $(x - 3)^2 = 0 \;\Rightarrow\; x = 3.$ There is a single (repeated) root, $x = 3$. A student who "expects two answers" might report $\pm 3$, but $-3$ does not satisfy the original equation: $(-3)^2 = 9$ while $6(-3) - 9 = -27$.

---

Example 2: System, Asking for a Product (Medium)

If $x + y = 7$ and $2x - y = 5$, what is the value of $xy$?

Solution. Add the equations to eliminate $y$: $3x = 12$, so $x = 4$. Substitute: $4 + y = 7$, so $y = 3$. Then $xy = 4 \cdot 3 = 12$.

---

Example 3: QC with an Unconstrained Variable (Hard)

Quantity A: $x^2$ Quantity B: $x$ (no information about $x$)

Solution. Test critical values.

• $x = 2$: A $= 4$, B $= 2$ ⇒ A greater.
• $x = \tfrac12$: A $= 0.25$, B $= 0.5$ ⇒ B greater.

Because the ordering changes, the answer is (D). This is the classic QC trap: an unconstrained variable that behaves differently for fractions and for integers greater than 1.

---

Example 4: Absolute-Value Inequality (Medium)

Which of the following describes all values of $x$ for which $|2x - 3| < 7$?

Solution. Rewrite as a compound inequality: $-7 < 2x - 3 < 7.$ Add 3 throughout: $-4 < 2x < 10$. Divide by 2: $-2 < x < 5$. The solution set is the open interval $(-2, 5)$.

---

Example 5: Function Evaluation with a Negative Input (Medium)

If $f(x) = 2x^2 - 3x + 1$, what is $f(-2)$?

Solution. Substitute carefully, squaring before applying signs: $f(-2) = 2(-2)^2 - 3(-2) + 1 = 2(4) + 6 + 1 = 8 + 6 + 1 = 15.$ The common wrong answer 3 comes from mishandling $-3(-2)$ as $-6$ instead of $+6$.

---

Example 6: Line Through Two Points (Medium–Hard)

A line in the $xy$-plane passes through the points $(2, 1)$ and $(6, 9)$. What is the $y$-coordinate of the point where the line crosses the $y$-axis?

Solution. Slope: $m = \dfrac{9 - 1}{6 - 2} = \dfrac{8}{4} = 2$. Use point $(2, 1)$ in $y = mx + b$: $1 = 2(2) + b \;\Rightarrow\; b = 1 - 4 = -3.$ The line is $y = 2x - 3$, so it crosses the $y$-axis at $y = -3$.