3.2 Quadratic Equations & Systems of Equations

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.

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1. Standard Form and Factoring Quadratics

A quadratic equation is a second-degree polynomial equation. Its standard form is:

$$ax^2 + bx + c = 0 \quad (a \neq 0)$$

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 $x = 2.5$.

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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:

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 $k$: $12k = 144 \implies k = 12$.

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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 $3$.

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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

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 $xy = 4 \times 3 = 12$.

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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.