3.2 Quadratic Equations & Systems of Equations
Key Takeaways
- Standard quadratic form is ax² + bx + c = 0; roots are found by factoring or the quadratic formula x = (-b ± √(b² - 4ac)) / (2a).
- The discriminant Δ = b² - 4ac determines root count: Δ > 0 gives 2 real roots, Δ = 0 gives 1 real root, and Δ < 0 gives 0 real roots.
- The vertex of parabola y = ax² + bx + c occurs at x = -b / (2a), representing the minimum when a > 0 and maximum when a < 0.
- A 2x2 system of linear equations has 1 solution if slopes differ, 0 solutions if lines are parallel, and infinitely many solutions if lines are identical.
Last updated: July 2026
3.2 Quadratic Equations & Systems of Equations
Quadratic equations and systems of equations are central components of GRE Quantitative Reasoning. Questions test factoring technique, root calculation, vertex identification, discriminant properties, and solving $2 \times 2$ systems of linear equations.
1. Standard Form and Factoring Quadratics
A quadratic equation is a second-degree polynomial equation. Its standard form is:
eq 0)$$ ### Factoring Quadratics ($a = 1$) To factor $x^2 + bx + c = 0$, find two numbers $m$ and $n$ such that: $$m \cdot n = c \quad \text{and} \quad m + n = b$$ Then the quadratic factors as $(x + m)(x + n) = 0$. By the Zero Product Property, the roots are $x = -m$ and $x = -n$. ### Factoring by Grouping / AC Method ($a eq 1$) 1. Find two numbers $m$ and $n$ such that $m \cdot n = a \cdot c$ and $m + n = b$. 2. Rewrite the middle term $bx$ as $mx + nx$. 3. Factor by grouping the first two terms and last two terms. #### Worked Example 3.2.1: Factoring when $a eq 1$ Solve for $x$: $2x^2 - 7x + 5 = 0$. **Solution:** 1. Identify $a = 2, b = -7, c = 5$. Compute $a \cdot c = 2 \times 5 = 10$. 2. Find two numbers that multiply to $10$ and sum to $-7$: the numbers are $-2$ and $-5$. 3. Rewrite middle term: $2x^2 - 2x - 5x + 5 = 0$. 4. Factor by grouping: $2x(x - 1) - 5(x - 1) = 0 \implies (2x - 5)(x - 1) = 0$. 5. Apply Zero Product Property: $2x - 5 = 0 \implies x = 2.5$; $x - 1 = 0 \implies x = 1$. The roots are $x = 1$ and $x = 2.5$. --- ## 2. The Quadratic Formula and the Discriminant When a quadratic equation cannot be easily factored, use the **Quadratic Formula**: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ ### The Discriminant ($\Delta = b^2 - 4ac$) The expression under the radical sign, $b^2 - 4ac$, is called the **discriminant** ($\Delta$). It determines the number and type of real roots: | Discriminant Value ($\Delta$) | Nature of Roots | Graph Intersection with x-axis | | :--- | :--- | :--- | | $\Delta > 0$ | **2 distinct real roots** | Parabola crosses the x-axis at 2 points | | $\Delta = 0$ | **1 real root (double root)** | Parabola touches (is tangent to) the x-axis at 1 point | | $\Delta < 0$ | **0 real roots** (2 complex roots) | Parabola does not touch or cross the x-axis | #### Worked Example 3.2.2: Discriminant Analysis Determine the value of $k$ for which the equation $3x^2 - 12x + k = 0$ has exactly one real solution. **Solution:** 1. For exactly one real solution, the discriminant must equal zero: $\Delta = b^2 - 4ac = 0$. 2. Identify coefficients: $a = 3, b = -12, c = k$. 3. Substitute into discriminant: $(-12)^2 - 4(3)(k) = 0 \implies 144 - 12k = 0$. 4. Solve for $k$: $12k = 144 \implies k = 12$. --- ## 3. Parabola Geometry & Vertex Coordinates The graph of a quadratic function $f(x) = ax^2 + bx + c$ is a **parabola**. - If $a > 0$, the parabola opens **upward** ($\cup$), and the vertex is the **minimum point**. - If $a < 0$, the parabola opens **downward** ($\cap$), and the vertex is the **maximum point**. ### Axis of Symmetry and Vertex Formula - **Axis of Symmetry (x-coordinate of vertex):** $$x = -\frac{b}{2a}$$ - **Vertex Coordinates:** $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$. #### Worked Example 3.2.3: Finding Parabola Vertex Minimum Find the minimum value of the quadratic function $f(x) = 2x^2 - 8x + 11$. **Solution:** 1. Identify coefficients: $a = 2, b = -8, c = 11$. Since $a = 2 > 0$, the parabola opens upward and has a minimum at its vertex. 2. Find the x-coordinate of the vertex: $x = -\frac{-8}{2(2)} = \frac{8}{4} = 2$. 3. Evaluate $f(2)$ to find the minimum y-value: $$f(2) = 2(2)^2 - 8(2) + 11 = 2(4) - 16 + 11 = 8 - 16 + 11 = 3$$ The minimum y-value of the function is $3$. --- ## 4. Systems of $2 \times 2$ Linear Equations A $2 \times 2$ system consists of two linear equations with two unknown variables: $$\begin{cases} a_1 x + b_1 y = c_1 \ a_2 x + b_2 y = c_2 \end{cases}$$ ### Solving Methods 1. **Substitution Method:** Isolate one variable in one equation and substitute its algebraic equivalent into the second equation. Best when one variable has a coefficient of $\pm 1$. 2. **Elimination Method:** Multiply one or both equations by constants so that adding or subtracting the equations cancels out one variable. Best when coefficients can easily match. ### Number of Solutions & Graphical Meaning | System Type | Graph Relationship | Number of Solutions | Condition on Slopes & Intercepts | | :--- | :--- | :--- | :--- | | **Independent / Consistent** | Intersecting lines | **Exactly 1 solution** | Different slopes ($m_1 eq m_2$) | | **Inconsistent** | Parallel lines | **0 solutions (No solution)** | Equal slopes ($m_1 = m_2$), different y-intercepts ($b_1 eq b_2$) | | **Dependent / Consistent** | Coincident (identical) lines | **Infinitely many solutions** | Equal slopes ($m_1 = m_2$) AND equal y-intercepts ($b_1 = b_2$) | #### Worked Example 3.2.4: Solving a $2 \times 2$ System by Elimination Solve the system of equations: $$\begin{cases} 3x + 2y = 18 \ 5x - 2y = 14 \end{cases}$$ **Solution:** 1. Notice that the coefficients of $y$ are $+2$ and $-2$. Add the two equations directly to eliminate $y$: $$(3x + 5x) + (2y - 2y) = 18 + 14 \implies 8x = 32 \implies x = 4$$ 2. Substitute $x = 4$ into the first equation: $$3(4) + 2y = 18 \implies 12 + 2y = 18 \implies 2y = 6 \implies y = 3$$ 3. The solution point is $(x, y) = (4, 3)$. Product $xy = 4 \times 3 = 12$. --- ## 5. GRE Quantitative Comparison Traps in Quadratics - **Two Roots Trap:** Remember that $x^2 = 25$ has TWO solutions: $x = 5$ and $x = -5$. However, the principal square root symbol $\sqrt{25}$ equals ONLY $+5$. Do not confuse $x^2 = k$ with $x = \sqrt{k}$. - **Assuming Both Variables are Positive:** In systems of equations or quadratic quantitative comparison questions, double-check whether variables are specified as positive, negative, or unconstrained.Test Your Knowledge
What are the roots of the quadratic equation x² - 5x - 24 = 0?
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Test Your Knowledge
For what value of k does the quadratic equation 3x² - 12x + k = 0 have exactly one real solution?
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Test Your Knowledge
What is the minimum y-value of the quadratic function f(x) = 2x² - 8x + 11?
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If 3x + 2y = 18 and 5x - 2y = 14, what is the value of the product xy?
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