Quadratic Equations · MathPrisma

Zero is the only number that confesses

Suppose I tell you two hidden numbers multiply to $12$ . What are they? Could be $3$ and $4$ . Could be $2$ and $6$ , or $-1$ and $-12$ , or $\tfrac12$ and $24$ — you know almost nothing. Now suppose I tell you they multiply to $0$ . Suddenly you know something for certain: one of them is zero. There is no other way to build a zero product. Twelve keeps secrets; zero confesses.

That one confession is why M2 taught you to factor. An equation like $x^2 + 5x + 6 = 0$ asks a genuinely hard question — which $x$ makes this whole expression collapse to nothing? — and factored form answers it almost by accident.

Zero-product: factoring finally gets paid

factor

M2’s hunt: multiply to

6

, add to

5

— it’s

2

and

3

. So the equation says

(x+2)(x+3) = 0

confess

A product of two things is zero only when one of them is:

x + 2 = 0

x + 3 = 0

solve each

Two one-step equations from A1:

x = -2

x = -3

check

Plug

-2

back in:

4 - 10 + 6 = 0

✓. Two answers — quadratics usually keep a pair.

The SAT often skips the factoring and hands you the confession directly: $(x - 3)(x + 5) = 0$ , solve. Read the roots straight off — each parenthesis dies at the value that zeroes it: $x = 3$ and $x = -5$ . Note both signs flipped. The factor $(x - 3)$ has root $+3$ ; answering $-3$ is the wrong answer the test is fishing for.

When the hunt comes up dry

M2 left you with a promise about $x^2 + 6x + 2$ : no integer pair multiplies to $2$ and adds to $6$ , but completing the square still rewrites it as $(x+3)^2 - 7$ . Watch what that buys you the moment there’s an equals sign:

$x^2 + 6x + 2 = 0 \;\;\Rightarrow\;\; (x+3)^2 - 7 = 0 \;\;\Rightarrow\;\; (x+3)^2 = 7$

Now the question is one you finished training for in F7: what squares to $7$ ? Two numbers do — $\sqrt{7}$ and $-\sqrt{7}$ . That’s where the famous $\pm$ is born: not from the $\sqrt{\phantom{7}}$ symbol (which always means the positive root), but from the equation, because both signs square to the same thing:

$x + 3 = \pm\sqrt{7} \;\;\Rightarrow\;\; x = -3 \pm \sqrt{7}$

Two exact answers, no factoring required. And nothing about this move needed the numbers to be friendly — completing the square solves any quadratic. Which raises a beautifully lazy question: if the recipe always works, why do we keep running it from scratch?

The formula: completing the square, done once, forever

Run the exact same recipe on $ax^2 + bx + c = 0$ — letters left in, nothing else new — and out comes the recipe’s own answer, prepackaged:

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

That’s all the quadratic formula is: completing the square, performed once in general so nobody ever has to do it again. Watch the whole build happen at the board — every move named as it is made, from $ax^2 + bx + c = 0$ to the boxed formula:

Transcript

You've probably seen this formula before. x equals negative b, plus or minus the square root of b squared minus four a c, all over two a. Most people memorize it without ever knowing where it comes from. Today we're going to build it, step by step — because once you've watched it come together, you don't have to trust it. You'll know why it works.

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

Here's our starting point: a x squared, plus b x, plus c, equals zero. The letters a, b, and c stand for any numbers you like. That's the whole idea — if we can solve this equation once, with letters, we've solved every quadratic equation at the same time. Our tool is a technique called completing the square. If you've met it before, this will feel familiar — and if you haven't, no problem at all: we're going to do every single step together.

$a x^{2} + b x + c = 0$

Step one. Completing the square works best when x squared stands alone, with no coefficient in front. Right now it has an a. So we divide both sides — every single term — by a. a x squared divided by a is x squared. b x divided by a is b over a, times x. c divided by a is c over a. And zero divided by a is still zero. Same equation, same solutions — just easier to work with.

$x^{2} + \frac{b}{a} x + \frac{c}{a} = 0$

We're ready to complete the square, and the recipe starts with the coefficient in front of x — here, that's b over a. Take half of it. Half of b over a is b over two a. Keep an eye on that expression — it's about to do all the work.

$\frac{1}{2} \cdot \frac{b}{a} = \frac{b}{2 a}$

The second step is to square that half: b over two a, squared, is b squared over four a squared. Here's why we want it: if we add this exact amount, the first terms become a perfect square. But we can't just add something to one side of an equation — that would change the equation. So we do it honestly: add it, and subtract it, in the same line. Adding and subtracting the same amount changes nothing at all.

$(\frac{b}{2 a})^{2} = \frac{b ^{2}}{4 a ^{2}}$

$x^{2} + \frac{b}{a} x + \frac{b ^{2}}{4 a ^{2}} - \frac{b ^{2}}{4 a ^{2}} + \frac{c}{a} = 0$

The payoff is sitting in the first three terms: x squared, plus b over a x, plus b squared over four a squared. Together, they are exactly x plus b over two a, squared. If you're not sure, multiply it out and check — square the first, two times the product, square the last. The left side has become a perfect square, with two leftover terms trailing behind it.

$(x + \frac{b}{2 a})^{2} - \frac{b ^{2}}{4 a ^{2}} + \frac{c}{a} = 0$

$(x + k)^{2} = x^{2} + 2 k x + k^{2}$

Time to tidy up. The two leftovers — minus b squared over four a squared, and plus c over a — don't belong with the square. Move each one to the right side; each flips its sign as it crosses the equals sign. Now we have: x plus b over two a, squared, equals b squared over four a squared, minus c over a.

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

The right side is two fractions, so let's combine them into one. They need a common denominator. Multiply c over a by four a over four a — that's multiplying by one, so its value doesn't change — and it becomes four a c over four a squared. Now subtract the numerators: b squared minus four a c, all over four a squared. Take a good look at that top — b squared minus four a c. It's important enough to have its own name, and it can tell you how many solutions an equation has before you even solve it. We'll come back to it.

$\frac{c}{a} \cdot \frac{4 a}{4 a} = \frac{4 a c}{4 a ^{2}}$

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

Now we take the square root of both sides — and this is where most mistakes happen, so let's make the mistake on purpose. You might write: x plus b over two a equals the square root of b squared minus four a c, over two a. It looks finished. But remember: if something squared equals nine, that something could be three — or negative three. Both work. A square has two square roots, one positive and one negative, and this line only kept one of them. It's missing half the answers — so it comes off the board.

$x + \frac{b}{2 a} = \frac{b ^{2} - 4 a c}{2 a}$

$3^{2} = 9, (- 3)^{2} = 9$

The correct line carries a plus-or-minus sign: x plus b over two a equals plus or minus the square root of b squared minus four a c, over two a. One small symbol — and it's the difference between finding both solutions and silently losing one.

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

We have one step left, and it's the easiest one: get x by itself. Subtract b over two a from both sides. Both pieces already share the denominator two a, so they combine into a single fraction. And there it is: x equals negative b, plus or minus the square root of b squared minus four a c, all over two a. That's the quadratic formula — and this time you didn't memorize it, you built it.

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

Step back and look at what we actually did. We took the general equation and completed the square, once. That's the whole secret — the formula is completing the square, done in advance, for every equation at the same time. The plus-or-minus is the square root keeping both answers. And b squared minus four a c, sitting under the root, is the part that decides how many solutions there are — more on that soon. From here on, solving a quadratic is reading off a, b, and c.

The quadratic formula, built from scratch · 6:35

Every piece has a birthplace — the $-b$ and $2a$ come from the “half of the middle” fold, and $b^2 - 4ac$ is the square’s leftover. Use it on the trinomial M2 proved unfactorable, $x^2 + 3x + 1 = 0$ : here $a = 1$ , $b = 3$ , $c = 1$ , so

$x = \dfrac{-3 \pm \sqrt{3^2 - 4(1)(1)}}{2(1)} = \dfrac{-3 \pm \sqrt{5}}{2}$

The dead end from M2’s slider — cracked in two lines.

Simplify the answer — F7 was training for this exact moment

Solve $x^2 + 4x - 8 = 0$ . The formula gives $b^2 - 4ac = 16 + 32 = 48$ , so

$x = \dfrac{-4 \pm \sqrt{48}}{2}$

F7 taught you to never leave $\sqrt{48}$ unopened — it deliberately stopped at $\sqrt{48} = 4\sqrt{3}$ and called the next move M3’s. This is that move:

$x = \dfrac{-4 \pm 4\sqrt{3}}{2} = \dfrac{2\left(-2 \pm 2\sqrt{3}\right)}{2} = -2 \pm 2\sqrt{3}$

Look hard at the middle step. Cancelling a fraction divides factors of the whole top — and $-4 \pm 4\sqrt{3}$ is a sum, not a product, so you must factor the $2$ out of every term before it may cancel. The tempting shortcut — cancel the $-4$ against the $2$ , leave the $4\sqrt{3}$ untouched, write $-2 \pm 4\sqrt{3}$ — feels natural precisely because it works on single-term fractions. Here it changes the answer, and the SAT lists it among the choices every time.

Try the move yourself — pick divisors and watch which terms cooperate:

Reduce the formula's answer by hand

Predict before you click: on the $x^2+4x-8$ preset, will ÷3 go through? Decide which term refuses, then click it and see. Then load $4x^2 - 4x - 1$ — the top and bottom share a $4$ : does it matter whether you take it in one ÷4 bite or two ÷2 bites? And $x^2 + 3x + 1$ ends somewhere important: with nothing to divide. $\tfrac{-3 \pm \sqrt{5}}{2}$ is finished — a denominator is not dirt, and recognizing “done” is worth points.

The discriminant: count the answers without finding them

Here’s the SAT’s favorite twist: “How many real solutions does $2x^2 - 5x + 7 = 0$ have?” — and solving it would be falling for the trap. Look at the formula’s anatomy. The $-b$ and the $2a$ are fixed scaffolding; the only part that can change how many answers exist is the piece under the root, $b^2 - 4ac$ , the discriminant:

$b^2 - 4ac > 0$ : the root is a real, positive number — the $\pm$ spreads two different answers.
$b^2 - 4ac = 0$ : the root is $0$ — the $\pm$ adds and subtracts nothing — exactly one answer.
$b^2 - 4ac < 0$ : you’d need $\sqrt{\text{negative}}$ , and F7 said it straight: no real number squares to a negative. No real solution — that’s a complete, correct answer, not a failure.

So: $(-5)^2 - 4(2)(7) = 25 - 56 = -31 < 0$ . Zero real solutions, ten seconds, no solving. (When you meet parabolas in M4, you’ll see these three cases as a curve crossing, touching, or missing the $x$ -axis. For now the number line tells the whole story.)

Watch the answers appear and vanish

Predict before you drag: the opening preset is $x^2 + 2x - 3 = 0$ , twins at $-3$ and $1$ . Drag $c$ upward one step at a time — each step drains the discriminant by $4$ . At what $c$ do the twins kiss? Check with algebra: $b^2 - 4ac = 4 - 4c = 0$ at $c = 1$ . That drag was the SAT’s “find $k$ so the equation has exactly one solution” question — your hand solved $b^2 - 4ac = 0$ before your pencil did. Then load $x^2 - 6x + 7$ and read the twins’ exact labels: $3 \pm \sqrt{2}$ , the formula’s answers living on the line. And notice, through every drag: the twins never leave their balance point $-\tfrac{b}{2a}$ — the $\pm$ spreads them evenly around it.

Sum and product: answers about the answers

That symmetry has a cash value. Add the formula’s two answers and the $\pm$ halves kill each other:

$\dfrac{-b + \sqrt{D}}{2a} + \dfrac{-b - \sqrt{D}}{2a} = \dfrac{-2b}{2a} = -\dfrac{b}{a}$

So when the SAT asks “what is the sum of the solutions of $2x^2 - 8x + 3 = 0$ ?”, it is testing whether you’ll spend ninety seconds solving — or read $-\tfrac{b}{a} = \tfrac{8}{2} = 4$ off the coefficients. (The product collapses the same way, to $\tfrac{c}{a}$ .) The answers were never hiding; they were encoded in the equation the whole time — the same lesson the pair hunt taught in M2, one floor up.

The one thing to remember

Get zero alone on one side — only zero confesses. If the left side factors, each factor hands you a root. If it doesn’t, the formula is completing the square already done for you: read $a$ , $b$ , $c$ with their signs, simplify the root the F7 way, and reduce every term or none. And when the question is only how many, don’t solve — the discriminant $b^2 - 4ac$ knew before you did.

The decision tree

See	Do	Example
factored $= 0$	read each root (flip the sign)	$(x-3)(x+5)=0 \to x = 3,\, -5$
it factors	zero-product: set each factor $= 0$	$x^2+5x+6=0 \to x = -2,\, -3$
$(x+h)^2 = k$	square root both sides, $\pm$ appears	$(x+3)^2 = 7 \to x = -3 \pm \sqrt{7}$
nothing factors	the quadratic formula	$x^2+3x+1=0 \to x = \tfrac{-3 \pm \sqrt{5}}{2}$
”how many solutions?“	discriminant only — don’t solve	$b^2 - 4ac$ , read the sign
”sum/product of solutions?”	Vieta — don’t solve	sum $-\tfrac{b}{a}$ , product $\tfrac{c}{a}$

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

The discriminant $D = b^2 - 4ac$ : $\;D > 0$ → two real solutions. $\;D = 0$ → exactly one. $\;D < 0$ → no real solution. “Exactly one solution” in a question is code for set $b^2 - 4ac = 0$ and solve for the unknown coefficient.

Simplest radical form: simplify the root ( $\sqrt{48} = 4\sqrt{3}$ ), then divide all three of $-b$ , the root’s coefficient, and $2a$ by their common factor.

Traps that cost points

Zero-product needs a zero: $(x-2)(x-5)=4$ never splits into $x-2=4$ or $x-5=4$ . Move the $4$ over and refactor.
Factored roots flip the sign: $(x - 3) = 0$ at $x = 3$ , not $-3$ .
In the formula, substitute with parentheses: $b = -5$ gives $-b = 5$ and $b^2 = 25$ — a $b^2$ is never negative.
The half-cancel: $\tfrac{-4 \pm 4\sqrt{3}}{2} \ne -2 \pm 4\sqrt{3}$ . Factor the top first; divide every term or none.
The $\pm$ comes from the equation ( $x^2 = 7$ has two answers), not from the $\sqrt{\phantom{x}}$ symbol — $\sqrt{7}$ alone is one positive number (F7’s rule).
“No real solution” is an answer. If $b^2 - 4ac < 0$ , say so and stop — hunting for arithmetic errors there is the trap.