Deriving the Quadratic Formula

Completing the square, done once with letters

The quadratic formula says that the solutions of $ax^{2} + bx + c = 0$ are $x = \frac{{-b \pm \sqrt{b^{2} - 4ac}}}{2a}$. It comes from completing the square on the general equation, with the letters a, b and c in place of numbers.

Each step below is made twice: first on the equation $2x^{2} + 8x - 10 = 0$, then on $ax^{2} + bx + c = 0$. The letters go through exactly the same moves as the numbers.

Make $x^{2}$ stand alone

Completing the square is easiest when $x^{2}$ has no number in front of it, so divide every term by the leading coefficient. With numbers: dividing $2x^{2} + 8x - 10 = 0$ by 2 gives $x^{2} + 4x - 5 = 0$.

With letters: dividing $ax^{2} + bx + c = 0$ by a gives $x^{2} + \left(\frac{b}{a}\right)x + \frac{c}{a} = 0$. Dividing by a is allowed because a is not 0: if a were 0, there would be no $x^{2}$ term, and the equation would not be a quadratic at all.

Move the constant across

Subtract the constant term from both sides, so that only the $x^{2}$ and x terms are left on the left. With numbers: $x^{2} + 4x = 5$. With letters: $x^{2} + \left(\frac{b}{a}\right)x = - \frac{c}{a}$.

Complete the square

Recall that $\left(x + p\right)^{2} = x^{2} + 2px + p^{2}$. The coefficient of x is 2p, so p is half the coefficient of x. With numbers: half of 4 is 2, and $\left(x + 2\right)^{2} = x^{2} + 4x + 4$. That is $x^{2} + 4x$ with an extra 4, so add 4 to both sides of $x^{2} + 4x = 5$: $\left(x + 2\right)^{2} = 9$.

With letters, the coefficient of x is $\frac{b}{a}$, and half of it is $\frac{b}{2a}$. Squaring the bracket gives $\left(x + \frac{b}{2a}\right)^{2} = x^{2} + \left(\frac{b}{a}\right)x + \frac{b^{2}}{4a^{2}}$, because $\left(\frac{b}{2a}\right)^{2} = \frac{b^{2}}{4a^{2}}$. So add $\frac{b^{2}}{4a^{2}}$ to both sides of $x^{2} + \left(\frac{b}{a}\right)x = - \frac{c}{a}$: $\left(x + \frac{b}{2a}\right)^{2} = \frac{b^{2}}{4a^{2}} - \frac{c}{a}$.

Whatever is added to the left side must be added to the right side too, or the two sides are no longer equal.

The x term is split into two equal strips, one along each of two sides of the $x^{2}$ square. The corner that is still missing is a square whose side is half the coefficient of x. Drag the edge until the x term is 4x, as in $x^{2} + 4x = 5$: the corner is $2^{2} = 4$, the number added to both sides. For $x^{2} + \left(\frac{b}{a}\right)x$ the corner is $\left(\frac{b}{2a}\right)^{2} = \frac{b^{2}}{4a^{2}}$.

Put the right side over one denominator

The right side, $\frac{b^{2}}{4a^{2}} - \frac{c}{a}$, is two fractions with different denominators. Their common denominator is $4a^{2}$: multiply the top and bottom of $\frac{c}{a}$ by 4a to get $\frac{4ac}{4a^{2}}$. Then $\frac{b^{2}}{4a^{2}} - \frac{4ac}{4a^{2}} = \frac{{b^{2} - 4ac}}{4a^{2}}$, so the equation is now $\left(x + \frac{b}{2a}\right)^{2} = \frac{{b^{2} - 4ac}}{4a^{2}}$.

The top of that fraction, $b^{2} - 4ac$, is the discriminant. With numbers, a = 2, b = 8 and c = −10, so $b^{2} - 4ac = 64 - 4 \times 2 \times \left(-10\right) = 64 + 80 = 144$, and $4a^{2} = 4 \times 4 = 16$. The right side is $\frac{144}{16} = 9$, which matches $\left(x + 2\right)^{2} = 9$.

Take the square root of both sides

Two numbers square to 9: $3^{2} = 9$ and $\left(-3\right)^{2} = 9$. So when you take the square root of both sides, write $\pm$ in front. With numbers: $\left(x + 2\right)^{2} = 9$ gives $x + 2 = \pm3$.

With letters, the square root of the top is $\sqrt{b^{2} - 4ac}$, and the square root of the bottom, $4a^{2}$, is 2a. So $x + \frac{b}{2a} = \pm \frac{\sqrt{b^{2} - 4ac}}{2a}$. Strictly, $\sqrt{4a^{2}}$ is 2a only when a is positive; when a is negative it is −2a. The $\pm$ already covers both signs, so writing 2a loses nothing.

This step also shows why the discriminant decides how many real solutions there are. If $b^{2} - 4ac$ is negative, it has no real square root, and the equation has no real solutions. If it is 0, adding 0 and subtracting 0 give the same x, so there is one repeated root.

Move $\frac{b}{2a}$ across

Subtract $\frac{b}{2a}$ from both sides: $x = - \frac{b}{2a} \pm \frac{\sqrt{b^{2} - 4ac}}{2a}$. Both parts are over 2a, so write them as one fraction: $x = \frac{{-b \pm \sqrt{b^{2} - 4ac}}}{2a}$. That is the quadratic formula.

With numbers: $x + 2 = \pm3$ gives x = −2 + 3 = 1 or x = −2 − 3 = −5. The formula gives the same answers. With a = 2, b = 8 and c = −10, the discriminant is 144 and $\sqrt{144} = 12$, so $x = \frac{{-8 \pm 12}}{4}$. That is $\frac{4}{4} = 1$ or $- \frac{20}{4} = -5$.

Because every step was made on the letters, the formula works for every quadratic equation. You never have to complete the square again to solve one; the formula has done it already.

The usual mistakes

Adding $\frac{b^{2}}{4a^{2}}$ to the left side only. The bracket $\left(x + \frac{b}{2a}\right)^{2}$ brings in $\frac{b^{2}}{4a^{2}}$, so the same amount must be added to the right side.

Leaving out the $\pm$. Taking the square root of both sides gives two possibilities, + and −, and dropping one loses a solution.

Putting only the square root over 2a, as in $-b \pm \frac{\sqrt{b^{2} - 4ac}}{2a}$. The −b is divided by 2a as well, so the whole top, $-b \pm \sqrt{b^{2} - 4ac}$, sits over 2a.

Writing an area as an equation

The application below starts from a sheet of paper with a margin of the same width, x cm, all the way around the printing. The margin takes x off each end of every side, so each side of the printing is 2x shorter than the side of the sheet. On a sheet 60 cm by 40 cm, the printing is (60 − 2x) cm by (40 − 2x) cm.

Its area is length times width, (60 − 2x)(40 − 2x). Setting that equal to the area the printing must have gives an equation, and expanding the brackets turns it into a quadratic equation. Part (b) then makes the same moves with the letters L, W and A in place of the numbers.