Quadratic Equations

What a Quadratic Equation Is

A quadratic equation is any equation that can be written in the standard form

$ax^2 + bx + c = 0, \qquad a \neq 0.$

The requirement $a \neq 0$ is what makes it quadratic rather than linear — the highest power of the variable is $2$. Because a quadratic curve (a parabola) can cross the $x$-axis twice, once, or not at all, a quadratic equation can have two, one, or zero real solutions. Almost every SAT Advanced Math question that mentions a squared variable is really asking you to pick the fastest of four solving tools and apply it cleanly.

Tool 1 — Factoring and the Zero-Product Property

Factoring is the first tool to try because it is the fastest when it works. The engine behind it is the zero-product property: if a product equals zero, at least one factor must be zero.

Solve $x^2 - 5x + 6 = 0$. Find two numbers that multiply to $+6$ and add to $-5$; those are $-2$ and $-3$:

$\begin{aligned} x^2 - 5x + 6 &= 0 \\ (x - 2)(x - 3) &= 0 \\ x = 2 \quad &\text{or} \quad x = 3. \end{aligned}$

The single most common mistake here is forgetting to move everything to one side first. Factoring only works once the equation reads "$\text{(something)} = 0$." If a problem gives $x^2 = 5x - 6$, rewrite it as $x^2 - 5x + 6 = 0$ before you factor.

Tool 2 — The Square-Root Method

When there is no middle ($bx$) term, do not factor — just isolate the square and take the root of both sides, keeping both signs:

$\begin{aligned} 3x^2 - 48 &= 0 \\ x^2 &= 16 \\ x &= \pm 4. \end{aligned}$

The $\pm$ is essential: $x^2 = 16$ is true for both $x = 4$ and $x = -4$. This method also handles a squared binomial directly: $(x - 1)^2 = 9$ gives $x - 1 = \pm 3$, so $x = 4$ or $x = -2$.

Tool 3 — Completing the Square

Completing the square rewrites a quadratic as a perfect square plus a constant. Take half of the $x$-coefficient, square it, and add it to both sides:

$\begin{aligned} x^2 + 6x - 4 &= 0 \\ x^2 + 6x &= 4 \\ x^2 + 6x + 9 &= 4 + 9 \\ (x + 3)^2 &= 13 \\ x &= -3 \pm \sqrt{13}. \end{aligned}$

This is also how the vertex form of a parabola is produced, and it is the derivation that gives us the next tool.

Tool 4 — The Quadratic Formula

The quadratic formula solves every quadratic, factorable or not:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$

Read off $a$, $b$, and $c$ from standard form and substitute carefully — sign errors on $b$ and $c$ are the biggest hazard. For $x^2 - 4x - 1 = 0$, we have $a = 1$, $b = -4$, $c = -1$:

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-1)}}{2(1)} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}.$

Use the formula whenever the numbers do not factor nicely; it never fails.

The Discriminant — Counting Solutions Without Solving

The expression under the radical, $b^2 - 4ac$, is the discriminant, and its sign alone tells you how many real solutions exist:

Discriminant $b^2 - 4ac$	Real solutions	Graph meets $x$-axis
Positive	Two distinct real solutions	Crosses twice
Zero	One real solution (a double root)	Just touches (tangent)
Negative	No real solutions	Never touches

SAT problems love to ask "for what value of $k$ does this equation have exactly one real solution?" — that is a discriminant question in disguise. Set $b^2 - 4ac = 0$ and solve for the unknown constant. You never have to find the roots themselves.

A Note on Sum and Product of Roots

For $ax^2 + bx + c = 0$ with roots $r_1$ and $r_2$, the shortcuts

$r_1 + r_2 = -\frac{b}{a}, \qquad r_1 r_2 = \frac{c}{a}$

let you answer "what is the sum of the solutions?" instantly, without solving. If $2x^2 - 8x + 5 = 0$, the sum of the roots is $-(-8)/2 = 4$ — no factoring or formula needed.

Choosing a Method Under Time Pressure

• No middle term ($b = 0$) → square-root method.
• Factors are obvious small integers → factor and use the zero-product property.
• Nothing factors, or the numbers are ugly → quadratic formula.
• "How many solutions?" or "for what $k$…" → discriminant, do not solve.
• "Sum/product of the solutions?" → $-b/a$ and $c/a$.

Picking the right tool is worth as many points as the algebra itself.