Deriving the Quadratic Formula

Completing the square, done once with letters

The quadratic formula says that the solutions of $a{x}^2+bx+c=0$ax² + bx + c = 0 are $x=\frac{\left(-b\pm \sqrt{b^2-4ac}\right)}{2a}$x = (−b ± √(b² − 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$2x² + 8x − 10 = 0, then on $a{x}^2+bx+c=0$ax² + bx + c = 0. The letters go through exactly the same moves as the numbers.

Make $x^2$x² stand alone

Completing the square is easiest when $x^2$x² has no number in front of it, so divide every term by the leading coefficient. With numbers: dividing $2x^2+8x-10=0$2x² + 8x − 10 = 0 by 2 gives $x^2+4x-5=0$x² + 4x − 5 = 0.

With letters: dividing $a{x}^2+bx+c=0$ax² + bx + c = 0 by a gives $x^2+\left(\frac{b}{a}\right)x+\frac{c}{a}=0$x² + (b/a)x + c/a = 0. Dividing by a is allowed because a is not 0: if a were 0, there would be no $x^2$x² 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 $x^2$x² and x terms are left on the left. With numbers: $x^2+4x=5$x² + 4x = 5. With letters: $x^2+\left(\frac{b}{a}\right)x=-\frac{c}{a}$x² + (b/a)x = −c/a.

Complete the square

Recall that ${\left(x+p\right)}^2=x^2+2px+p^2$(x + p)² = x² + 2px + p². The coefficient of x is 2p, so p is half the coefficient of x. With numbers: half of 4 is 2, and ${\left(x+2\right)}^2=x^2+4x+4$(x + 2)² = x² + 4x + 4. That is $x^2+4x$x² + 4x with an extra 4, so add 4 to both sides of $x^2+4x=5$x² + 4x = 5: ${\left(x+2\right)}^2=9$(x + 2)² = 9.

With letters, the coefficient of x is $\frac{b}{a}$b/a, and half of it is $\frac{b}{2a}$b/2a. Squaring the bracket gives ${\left(x+\frac{b}{2a}\right)}^2=x^2+\left(\frac{b}{a}\right)x+\frac{b^2}{4a^2}$(x + b/2a)² = x² + (b/a)x + b²/4a², because ${\left(\frac{b}{2a}\right)}^2=\frac{b^2}{4a^2}$(b/2a)² = b²/4a². So add $\frac{b^2}{4a^2}$b²/4a² to both sides of $x^2+\left(\frac{b}{a}\right)x=-\frac{c}{a}$x² + (b/a)x = −c/a: ${\left(x+\frac{b}{2a}\right)}^2=\frac{b^2}{4a^2}-\frac{c}{a}$(x + b/2a)² = b²/4a² − 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 $x^2$x² 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 $x^2+4x=5$x² + 4x = 5: the corner is $2^2=4$2² = 4, the number added to both sides. For $x^2+\left(\frac{b}{a}\right)x$x² + (b/a)x the corner is ${\left(\frac{b}{2a}\right)}^2=\frac{b^2}{4a^2}$(b/2a)² = b²/4a².

Put the right side over one denominator

The right side, $\frac{b^2}{4a^2}-\frac{c}{a}$b²/4a² − c/a, is two fractions with different denominators. Their common denominator is $4a^2$4a²: multiply the top and bottom of $\frac{c}{a}$c/a by 4a to get $\frac{4ac}{4a^2}$4ac/4a². Then $\frac{b^2}{4a^2}-\frac{4ac}{4a^2}=\frac{\left(b^2-4ac\right)}{4a^2}$b²/4a² − 4ac/4a² = (b² − 4ac)/4a², so the equation is now ${\left(x+\frac{b}{2a}\right)}^2=\frac{\left(b^2-4ac\right)}{4a^2}$(x + b/2a)² = (b² − 4ac)/4a².

The top of that fraction, $b^2-4ac$b² − 4ac, is the discriminant. With numbers, a = 2, b = 8 and c = −10, so $b^2-4ac=64-4\times 2\times \left(-10\right)=64+80=144$b² − 4ac = 64 − 4 × 2 × (−10) = 64 + 80 = 144, and $4a^2=4\times 4=16$4a² = 4 × 4 = 16. The right side is $\frac{144}{16}=9$144/16 = 9, which matches ${\left(x+2\right)}^2=9$(x + 2)² = 9.

Take the square root of both sides

Two numbers square to 9: $3^2=9$3² = 9 and ${\left(-3\right)}^2=9$(−3)² = 9. So when you take the square root of both sides, write $\pm$± in front. With numbers: ${\left(x+2\right)}^2=9$(x + 2)² = 9 gives $x+2=\pm 3$x + 2 = ±3.

With letters, the square root of the top is $\sqrt{b^2-4ac}$√(b² − 4ac), and the square root of the bottom, $4a^2$4a², is 2a. So $x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}$x + b/2a = ±√(b² − 4ac)/2a. Strictly, $\sqrt{4a^2}$√(4a²) 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 $b^2-4ac$b² − 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 $\frac{b}{2a}$b/2a across

Subtract $\frac{b}{2a}$b/2a from both sides: $x=-\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}$x = −b/2a ± √(b² − 4ac)/2a. Both parts are over 2a, so write them as one fraction: $x=\frac{\left(-b\pm \sqrt{b^2-4ac}\right)}{2a}$x = (−b ± √(b² − 4ac))/2a. That is the quadratic formula.

With numbers: $x+2=\pm 3$x + 2 = ±3 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 $\sqrt{144}=12$√144 = 12, so $x=\frac{\left(-8\pm 12\right)}{4}$x = (−8 ± 12)/4. That is $\frac{4}{4}=1$4/4 = 1 or $-\frac{20}{4}=-5$−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 $\frac{b^2}{4a^2}$b²/4a² to the left side only. The bracket ${\left(x+\frac{b}{2a}\right)}^2$(x + b/2a)² brings in $\frac{b^2}{4a^2}$b²/4a², 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 $-b\pm \frac{\sqrt{b^2-4ac}}{2a}$−b ± √(b² − 4ac)/2a. The −b is divided by 2a as well, so the whole top, $-b\pm \sqrt{b^2-4ac}$−b ± √(b² − 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.