Quadratics and completing the square

Quadratics and parabolas

A quadratic is $y = ax^2 + bx + c$, where $a$, $b$ and $c$ are numbers and $a \neq 0$. With $a = 0$ it would be the line $y = bx + c$ of lesson 1.2.1. Its graph is a U-shaped curve called a parabola.

Here is $y = x^2 - 6x + 5$ at seven inputs.

The values fall, turn at $x = 3$, then climb back in mirror image. The turning point $(3, -4)$ is the vertex.

The vertical line $x = 3$ through it is the axis of symmetry. Inputs the same distance either side, such as 2 and 4, give the same output.

The vertex answers two questions at once: how small can the quadratic get, and at which input? For $x^2 - 6x + 5$ the smallest value is $-4$, at $x = 3$. In the words of lesson 1.1.1, its range is $[-4, \infty)$.

Three things can be read off the formula before any working.

• The sign of $a$. If $a > 0$ the parabola opens upwards and the vertex is its lowest point. If $a < 0$ it opens downwards and the vertex is its highest point.
• The size of $a$. As in lesson 1.1.2, $2x^2$ is a narrower stretch of $x^2$, and $\frac{1}{2}x^2$ a wider one.
• The value of $c$. At $x = 0$ the quadratic equals $c$, so $(0, c)$ is the $y$-intercept. Here it is $(0, 5)$.

Completing the square

A table finds the vertex only when it lands on an input you tried. Completing the square finds it exactly. Lesson 1.1.1 used it once, to write $x^2 - 2x$ as $(x - 1)^2 - 1$.

Completing the square rewrites a quadratic as $a(x - h)^2 + k$. That is $x^2$ stretched by $a$, moved $h$ right and $k$ up (lesson 1.1.2), so its vertex is $(h, k)$.

The method rests on one square: $(x - 3)^2 = x^2 - 6x + 9$. Its $x$ term is $-6x$, twice the $-3$ in the bracket. So halving the coefficient of $x$ gives the number to put in the bracket.

1. 1
   $$x^2 - 6x + 5$$

   Why?
2. 2
   $$(x - 3)^2 - 9 + 5$$

   Why?
3. 3
   $$(x - 3)^2 - 4$$

   Why?

A square is never negative, so $(x - 3)^2 - 4$ is never below $-4$. It equals $-4$ exactly when $x - 3 = 0$, at $x = 3$. So the vertex is $(3, -4)$, as the table showed.

When $a$ is not 1

For $2x^2 - 8x + 3$, take the 2 out of the two $x$ terms first: $2(x^2 - 4x) + 3$. Inside the bracket, half of $-4$ is $-2$, so $x^2 - 4x = (x - 2)^2 - 4$. Then multiply back out:

The vertex is $(2, -5)$. Note that the $-4$ inside the bracket was multiplied by 2.

When $a$ is negative

For $-x^2 + 4x + 5$, take out $-1$ to leave $-(x^2 - 4x) + 5$, then complete the square inside as before:

Now the square is subtracted, so the value is never above 9. The vertex $(2, 9)$ is a maximum, and the parabola opens downwards, as $a = -1$ says.

When half of $b$ is a fraction

For $x^2 + 3x + 1$, half of 3 is 1.5, and $(x + 1.5)^2 = x^2 + 3x + 2.25$. Taking 2.25 away and adding 1 gives $(x + 1.5)^2 - 1.25$. The bracket is $x + 1.5 = x - (-1.5)$, so the vertex is $(-1.5, -1.25)$, to the left of the $y$-axis.

The same steps work for any quadratic. Taking $a$ out leaves $x^2 + \frac{b}{a}x$ in the bracket, and half of $\frac{b}{a}$ is $\frac{b}{2a}$. So the square is $\left(x + \frac{b}{2a}\right)^2$, and the vertex is at $h = -\frac{b}{2a}$.

Roots

The roots of a quadratic are the solutions of $ax^2 + bx + c = 0$. They are where the parabola meets the $x$-axis: its zeros, from lesson 1.1.1. The completed square leads straight to them.

Example 1. $(x - 3)^2 - 4 = 0$ gives $(x - 3)^2 = 4$. Both 2 and $-2$ square to 4, so $x - 3 = \pm 2$. The roots are $x = 1$ and $x = 5$, the two zeros in the table.

The roots sit either side of the axis by the same distance: $3 - 2$ and $3 + 2$. Every quadratic with two roots has this mirror pattern.

Example 2. $2(x - 2)^2 - 5 = 0$ gives $(x - 2)^2 = 2.5$, so $x = 2 \pm \sqrt{2.5}$. Since $\sqrt{2.5} \approx 1.5811$, the roots are about 0.4189 and 3.5811. A table of whole numbers would never land on them.

The same steps, done once on $ax^2 + bx + c = 0$ with letters, give a formula for every quadratic.

For $2x^2 - 8x + 3$, the number under the root is $64 - 24 = 40$, so $x = \frac{8 \pm \sqrt{40}}{4}$. Since $\frac{\sqrt{40}}{4} \approx 1.5811$, that is $2 \pm 1.5811$ again.

Two roots, one or none

The number under the square root, $b^2 - 4ac$, is the discriminant. Its sign says how many real roots there are.

• Positive: two roots. For $x^2 - 6x + 5$ it is $36 - 20 = 16$, and the roots are 1 and 5.
• Zero: one root. $x^2 - 6x + 9$ is $(x - 3)^2$. Its discriminant is $36 - 36 = 0$, and the vertex $(3, 0)$ touches the axis at the single root 3.
• Negative: no real roots. $x^2 - 6x + 13$ is $(x - 3)^2 + 4$, never below 4. Its discriminant is $36 - 52 = -16$.

A negative discriminant would need the square root of a negative number, which has no real value (lesson 1.1.1). Lesson 1.7.2 gives such roots a meaning, as complex numbers.

Lesson 1.3.1 shows how each root gives a factor. Case study 1.2.4 uses the vertex to find the least risky mix of two assets.

Common mistakes

1. The vertex is at $-\frac{b}{2a}$, minus sign included. For $x^2 - 6x + 5$ that is $+3$. And $(x + 1.5)^2$ puts the vertex at $-1.5$.
2. Keep the balance. $(x - 3)^2$ brings in an extra 9, so take 9 away: $x^2 - 6x + 5 = (x - 3)^2 - 9 + 5$.
3. Take $a$ out first, then multiply back. $2x^2 - 8x + 3$ is $2(x - 2)^2 - 5$, not $2(x - 2)^2 - 1$.
4. Minimum or maximum depends on the sign of $a$. $-(x - 2)^2 + 9$ has a highest value, 9, and no lowest.
5. $(x - 3)^2 = 4$ has two solutions. Both $x - 3 = 2$ and $x - 3 = -2$ count.
6. Divide all of $-b \pm \sqrt{b^2 - 4ac}$ by $2a$, not just the square root.
7. A negative discriminant means no real roots. Do not drop its minus sign to force a square root.
8. With $a = 0$ there is no quadratic. The formulas would divide by zero; solve the line $bx + c = 0$ instead.