Quadratic equations — factoring, completing the square

The core insight: We're transforming $ax^2 + bx + c = 0$ into a product $(x - p)(x - q) = 0$ or a perfect square $(x - h)^2 = k$ because if a product equals zero, at least one factor must be zero. This turns a polynomial problem into simple linear equations.

Method 1: Factoring

How to Factor (Step-by-Step Derivation)

Case 1: Monic quadratics ($a = 1$, so we have $x^2 + bx + c = 0$)

We seek two numbers that:

• Multiply to give $c$ (the constant term)
• Add to give $b$ (the coefficient of $x$)

Why? Because $(x + p)(x + q) = x^2 + (p+q)x + pq$. Matching coefficients:

• Coefficient of $x$: $p + q = b$
• Constant term: $pq = c$

What we do: Find pairs of factors of $c$, test which pair sums to $b$.

Why this step? Each factor represents a "boundary" where the parabola crosses the x-axis. Setting each factor to zero isolates those crossing points.

Case 2: Non-monic quadratics ($a \neq 1$, so we have $ax^2 + bx + c = 0$)

AC Method (Derivation):

1. Multiply $a \cdot c$ to get a target product.
2. Find two numbers that multiply to $ac$ and add to $b$.
3. Split the middle term $bx$ using those two numbers.
4. Factor by grouping.

Why does this work? We're transforming the problem into a monic-like structure by absorbing the leading coefficient into the constant term temporarily.

Why this step? Grouping isolates a common factor, revealing the hidden factorization.

Method 2: Completing the Square

Derivation from First Principles

Goal: Rewrite $ax^2 + bx + c$ so the $x^2$ and $x$ terms combine into $(x + \text{something})^2$.

Step-by-step construction:

Start with $ax^2 + bx + c = 0$.

Step 1: Factor out $a$ from the first two terms (make the $x^2$ coefficient 1 inside): $a\left(x^2 + \frac{b}{a}x\right) + c = 0$

Why? A perfect square $(x + k)^2 = x^2 + 2kx + k^2$ requires the $x^2$ coefficient to be 1.

Step 2: Identify the number to complete the square. A perfect square $(x + k)^2$ has its middle term as $2k$. So: $2k = \frac{b}{a} \quad \Rightarrow \quad k = \frac{b}{2a}$

The missing constant term in the perfect square is $k^2 = \left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$.

Step 3: Add and subtract this term inside the parentheses: $a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right) + c = 0$

Why add and subtract? We're not changing the equation's value, just rearranging it.

Step 4: Recognize the perfect square and simplify: $a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\right] + c = 0$ $a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c = 0$ $a\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a} - c = \frac{b^2 - 4ac}{4a}$

Step 5: Divide both sides by $a$: $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$

Step 6: Take the square root of both sides: $x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$

Step 7: Solve for $x$: $x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Verification: Substitute back: $(-1)^2 + 6(-1) + 5 = 1 - 6 + 5 = 0$ ✓, and $(-5)^2 + 6(-5) + 5 = 25 - 30 + 5 = 0$ ✓.

Mistake 1: Forgetting to factor out $a$ first Student writes: $2x^2 + 8x + 6 = 2(x + 2)^2$ ✗

Why it feels right: The student factors out the 2, sees $2(x^2 + 4x + 3)$, takes half of 4 to get 2, and jumps straight to $2(x+2)^2$—forgetting that $(x+2)^2 = x^2 + 4x + 4$ introduces an extra $+4$ that must be compensated by subtracting it.

The fix: Always make the $x^2$ coefficient equal to 1 first, then add and subtract the completing term: $2(x^2 + 4x + 3) = 2[(x+2)^2 - 4 + 3] = 2[(x+2)^2 - 1]$. The leftover $-1$ matters!

Mistake 2: Sign errors when factoring Student factors $x^2 - 5x + 6$ as $(x - 2)(x - 3)$ but then writes solutions as $x = -2, x = -3$ ✗

Why it feels right: The factors contain $-2$ and $-3$, so the student thinks those are the answers directly.

The fix: Remember, we set each factor equal to zero: $x - 2 = 0 \Rightarrow x = +2$. The signs flip when solving. Ask: "What value makes each factor zero?"

Mistake 3: Only adding the completing term to one side Student writes: $x^2 + 6x = -5$, then $x^2 + 6x + 9 = -5$ ✗

Why it feels right: They correctly add the completing-the-square term to the left but forget the equation must stay balanced.

The fix: Whatever you add to the left, add to the right. $x^2 + 6x + 9 = -5 + 9 = 4$.

Factoring is like breaking a big LEGO structure into two smaller towers. If the whole thing is flat on the ground (equals zero), one of the towers must be flat (one of the factors must equal zero). We find which pieces make each tower collapse.

Completing the square is like rearranging puzzle pieces to make a perfect square picture. Once you see the square, you can easily measure how far from the center the edges are—and those distances tell you where the bridge touches the ground. It's a clever trick that works every single time, even when factoring is too hard!

Both methods are tools in your math toolbox. Factoring is faster when you can spot the pattern. Completing the square is your superpower when numbers get messy—it's how we discovered the quadratic formula!

Factoring (AC method): "Multiply $ac$, Find two numbers, Split middle term, Group, Factor out"

Connections

• 2.1.15-Linear-equations-and-inequalities — Factoring reduces quadratics to linear equations
• 2.1.17-Quadratic-formuland-discriminant — Completing the square derives the quadratic formula
• 2.3.4-Parabolas-and-vertex-form — Completing the square reveals vertex form $y = a(x-h)^2 + k$
• 3.2.1-Polynomials-factoring-techniques — Factoring extends to higher-degree polynomials
• 4.1.2-Solving-systems-graphically — Quadratic solutions are x-intercepts on a graph

Flashcards

What is a quadratic equation? :: An equation of the form $ax^2 + bx + c = 0$ where $a \neq 0$.

Concept Map

Hinglish (regional understanding)