The quadratic formula

A quadratic equation $ax^2 + bx + c = 0$ has the solutions

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

Every course hands this over to be used. Here is where it comes from: it is completing the square, carried out once with letters instead of numbers, so that the answer works for every quadratic at once.

Before you start

You need to be able to complete the square, because that is the whole method. If $x^2 + 6x + 2 = 0$ , you write the left side as $(x + 3)^2 - 7$ and solve from there. The only difference below is that the numbers are $a$ , $b$ and $c$ .

Predict first

Before reading on, predict: the formula divides by $2a$ , so it breaks when $a = 0$ . What kind of equation is $ax^2 + bx + c = 0$ when $a = 0$ ? Work that out first, then read the proof and see where it is assumed.

The proof

Start with the general quadratic equation and $a \neq 0$ :

ax^2 + bx + c = 0

Step 1: divide by $a$ . This is where $a \neq 0$ is needed — you cannot divide by zero. Dividing turns the leading coefficient into $1$ , which is what completing the square wants:

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

Step 2: move the constant to the other side.

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

Step 3: complete the square. Take half of the coefficient of $x$ , which is $\dfrac{b}{2a}$ , and square it: $\dfrac{b^2}{4a^2}$ . Add it to both sides:

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

The left side is now a perfect square, by construction:

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

Step 4: put the right side over one denominator. The common denominator is $4a^2$ , so multiply the second fraction by $\dfrac{4a}{4a}$ :

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

Step 5: take the square root of both sides. A square has two square roots, one positive and one negative, and that is where the $\pm$ comes from:

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

Step 6: subtract $\dfrac{b}{2a}$ from both sides. The two fractions already share the denominator $2a$ :

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

That is the formula.

What the proof tells you for free

The discriminant is a square root. The quantity $b^2 - 4ac$ sits under a square root, so its sign decides how many solutions there are: positive gives two, zero gives one, negative gives none. That is not an extra rule to learn — it is the shape of the answer.

Why the formula fails when $a = 0$ . Step 1 divides by $a$ . When $a = 0$ the equation is $bx + c = 0$ , which is a straight line with one solution, not a quadratic. The formula was never meant to apply, and it says so by breaking.

Why you can divide by $2a$ at the end. The proof assumes $a \neq 0$ throughout, so $2a \neq 0$ and the division is allowed.

Check your understanding

Why is the $\pm$ there? Answer in terms of the step it appears in, not by quoting the formula.
Use the formula on $2x^2 - 7x + 3 = 0$ and check one root by substituting it.
The proof completes the square by adding $\dfrac{b^2}{4a^2}$ to both sides. Where did that expression come from?

Answers

It appears when taking a square root in Step 5. Every positive number has two square roots, so the equation $\text{(something)}^2 = k$ has two solutions whenever $k > 0$ .
$a = 2$ , $b = -7$ , $c = 3$ , so $b^2 - 4ac = 49 - 24 = 25$ and $x = \dfrac{7 \pm 5}{4}$ , giving $x = 3$ or $x = \tfrac{1}{2}$ . Check: $2(3)^2 - 7(3) + 3 = 18 - 21 + 3 = 0$ .
The coefficient of $x$ is $\dfrac{b}{a}$ . Completing the square always uses half of that coefficient squared, and half of $\dfrac{b}{a}$ is $\dfrac{b}{2a}$ , whose square is $\dfrac{b^2}{4a^2}$ .

A common error

The formula applies to $ax^2 + bx + c = 0$ with the terms collected and in that order. If the equation arrives as $3 + 2x^2 = 7x$ , rearrange it to $2x^2 - 7x + 3 = 0$ before reading off $a$ , $b$ and $c$ — and take the sign of $b$ with the term, because a negative $b$ is the most common way to lose a mark.