Quadratic formula — derivation by completing the square

The standard quadratic equation

Why these letters? Convention: $a$ multiplies the squared term, $b$ the linear term, $c$ is the constant. The restriction $a \neq 0$ ensures we actually have an $x^2$ term.

Derivation from first principles

Step 1: Normalize the leading coefficient

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

Why this step? Completing the square is easiest when the coefficient of $x^2$ is 1. We divide everything by $a$: $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$

What did we do? Divided both sides by $a$ (legal since $a \neq 0$). Now the $x^2$ term has coefficient 1.

Step 2: Isolate the terms with $x$

Move the constant to the right: $x^2 + \frac{b}{a}x = -\frac{c}{a}$

Why? We need space on the left to add the "completing the square" term.

Step 3: Complete the square

Why this magic number? Because $(x + h)^2 = x^2 + 2hx + h^2$. If we want the middle term to be $kx$, we need $2h = k$, so $h = \frac{k}{2}$. Then the constant term must be $h^2 = \left(\frac{k}{2}\right)^2$.

In our case, $k = \frac{b}{a}$, so we add $\left(\frac{b}{2a}\right)^2$ to both sides: $x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$

What's happening? Left side is now a perfect square: $\left(x + \frac{b}{2a}\right)^2$.

Step 4: Recognize the perfect square

$\left(x + \frac{b}{2a}\right)^2 = -\frac{c}{a} + \frac{b^2}{4a^2}$

How do we know? Expand $\left(x + \frac{b}{2a}\right)^2 = x^2 + 2 \cdot x \cdot \frac{b}{2a} + \left(\frac{b}{2a}\right)^2 = x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}$. ✓

Step 5: Simplify the right side

Get a common denominator: $-\frac{c}{a} + \frac{b^2}{4a^2} = -\frac{4ac}{4a^2} + \frac{b^2}{4a^2} = \frac{b^2 - 4ac}{4a^2}$

Why combine? So we can take the square root cleanly.

Now we have: $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$

Step 6: Take square roots

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

Why $\pm$? Because if $y^2 = k$, then $y = \pm\sqrt{k}$ (two solutions).

Simplify the square root: $\sqrt{\frac{b^2 - 4ac}{4a^2}} = \frac{\sqrt{b^2 - 4ac}}{\sqrt{4a^2}} = \frac{\sqrt{b^2 - 4ac}}{2|a|}$

Since we typically write the formula with $2a$ (not $2|a|$), and the $\pm$ covers both signs: $x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$

Step 7: Solve for $x$

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

Combine the fractions:

What does this mean? For any quadratic $ax^2 + bx + c = 0$, plug in $a, b, c$, and you get the two solutions (or one repeated solution if the discriminant is zero).

The discriminant

Why? The discriminant is inside the square root. If it's negative, $\sqrt{\Delta}$ isn't real.

Worked examples

Identify coefficients: $a = 1$, $b = -5$, $c = 6$.

Apply formula: $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}$

Why this step? Substitute directly into the quadratic formula.

$x = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm \sqrt{1}}{2} = \frac{5 \pm 1}{2}$

Evaluate both solutions: $x = \frac{5 + 1}{2} = 3 \quad \text{or} \quad x = \frac{5 - 1}{2} = 2$

Check: $(3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$ ✓, and $(2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$ ✓

Coefficients: $a = 1$, $b = -6$, $c = 9$.

Discriminant: $\Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0$.

Why check discriminant? $\Delta = 0$ tells us there's exactly one solution (a repeated root).

Apply formula: $x = \frac{-(-6) \pm \sqrt{0}}{2(1)} = \frac{6 \pm 0}{2} = 3$

Notice: This is $(x - 3)^2 = 0$, which factors perfectly.

Coefficients: $a = 2$, $b = 7$, $c = -4$.

Apply formula: $x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-4)}}{2(2)}$

Why this step? Carefully handle the negative $c$ and the $2a$ in the denominator.

$x = \frac{-7 \pm \sqrt{49 + 32}}{4} = \frac{-7 \pm \sqrt{81}}{4} = \frac{-7 \pm 9}{4}$

Two solutions: $x = \frac{-7 + 9}{4} = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-7 - 9}{4} = \frac{-16}{4} = -4$

Check: $2\left(\frac{1}{2}\right)^2 + 7\left(\frac{1}{2}\right) - 4 = \frac{1}{2} + \frac{7}{2} - 4 = 4 - 4 = 0$ ✓

Coefficients: $a = 1$, $b = 2$, $c = 5$.

Discriminant: $\Delta = 2^2 - 4(1)(5) = 4 - 20 = -16 < 0$.

What does this mean? No real solutions. In complex numbers: $x = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i$

Common mistakes

Why it feels right: You see $-5$ and just copy it.

The fix: The formula is $-b$, not $b$. Here $b = -5$, so $-b = -(-5) = +5$. Always identify $b$ first, then negate it.

Why it feels right: The numerator gets all the attention; the denominator is easy to overlook.

The fix: Write the full formula structure first: $x = \frac{\text{numerator}}{2a}$. The ENTIRE numerator (including the $\pm$ part) is divided by $2a$, not just part of it.

Why it feels right: "You can't take the square root of a negative number" in real numbers.

The fix: Specify "no real solutions." There ARE solutions in the complex numbers: $\sqrt{-k} = i\sqrt{k}$. The quadratic always has two roots (counting multiplicity) in $\mathbb{C}$.

Why it feels right: Forgetting that multiplying by a negative $c$ introduces a sign change.

The fix: $-4ac = -4 \cdot 2 \cdot (-4) = -8 \cdot (-4) = +32$. Always compute $4ac$ first, then subtract it: $49 - (-32) = 49 + 32 = 81$.

Connections

• Completing the square technique
• Perfect square trinomials
• Discriminant and nature of roots
• Factoring quadratics
• Complex numbers and quadratic equations
• Parabola and its vertex form
• Vieta's formulas for sum and product of roots
• Quadratic inequalities

Or visually: the formula is a fraction with $2a$ on bottom, and on top you have "opposite of $b$" plus/minus a square root containing "$b$ squared minus four-$a$-$c$."

Recall Explain to a 12-year-old

Imagine you have a mystery number $x$, and someone tells you: "If you square it, multiply it by something, add another multiple of it, and add a constant, you get zero." That sounds complicated! But here's the magic trick:

We reshape the equation by completing the square—it's like rearranging puzzle pieces so they fit into a perfect square shape, like $(x + \text{something})^2$. Once we have that, we can "undo" the square by taking a square root, which is way easier than trying to guess $x$.

The quadratic formula is the final answer to this puzzle. You just plug in three numbers ($a$, $b$, $c$) from your equation, and it spits out the mystery number $x$. Sometimes you get two answers (the parabola crosses the $x$-axis twice), sometimes one (it just touches), and sometimes none in regular numbers (it never crosses, but you can use imaginary numbers).

The coolest part? This formula works for any quadratic, no matter how messy the numbers are!

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Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho beta, yahan pe core idea ye hai ki jab humein $ax^2 + bx + c = 0$ jaise quadratic equation solve karni ho, to problem ye hoti hai ki $x$ do jagah aata hai — ek baar squared aur ek baar linear form mein. Isko seedha solve karna mushkil hai. To hum ek smart trick use karte hain jise completing the square kehte hain. Iska matlab hai ki hum left side ko ek perfect square, matlab $(x + \text{kuch})^2$ ke form mein badal dete hain. Ek baar perfect square ban gaya, to hum dono taraf square root le sakte hain aur $x$ ko aaram se alag kar sakte hain. Bilkul jaise ek uljhi hui knot ko suljhaana — usko ek simple loop mein reshape kar do.

Ab step-by-step derivation ka intuition ye hai: pehle hum poori equation ko $a$ se divide karte hain taaki $x^2$ ka coefficient 1 ban jaaye (kyunki square banana tab easy hota hai). Phir constant ko right side bhej dete hain. Uske baad wo magic number $\left(\frac{b}{2a}\right)^2$ dono taraf add karte hain — ye number randomly nahi aata, ye isliye aata hai kyunki $(x+h)^2 = x^2 + 2hx + h^2$, aur middle term match karane ke liye $h = \frac{b}{2a}$ lena padta hai. Bas isi se pura left side perfect square ban jaata hai, aur square root lekar humein wo famous quadratic formula mil jaati hai: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Ye important isliye hai kyunki ek baar formula yaad ho gaya, to tumhe kisi bhi quadratic equation ke liye completing the square baar-baar karne ki zaroorat nahi — sirf $a, b, c$ plug karo aur answer aa jaayega. Aur us formula ke andar jo $b^2 - 4ac$ term hai, use discriminant kehte hain — ye batata hai ki roots real honge, repeated honge ya complex. Exam mein ye directly kaam aata hai, aur samajhne ke baad tumhe rote yaad karne ki tension bhi nahi rahegi, kyunki tumhe pata hoga ye formula kahaan se aayi.