Quadratics (Completing the Square, Discriminant)

Why Completing the Square?

At GCSE you factorised quadratics like $x^2 + 5x + 6 = (x+2)(x+3)$. But try factorising $x^2 + 6x + 1$. It doesn’t split into nice integer factors.

Completing the square is the technique that handles every quadratic — factorisable or not. It also reveals the vertex of the parabola, which is essential for sketching, and leads us naturally to the discriminant.

Key Concepts

Completing the square (a = 1)

The idea is to rewrite $x^2 + bx + c$ in the form $(x + p)^2 + q$, where the vertex of the parabola sits at $(-p,\, q)$.

Start with the $x^2 + bx$ part. We take half the coefficient of $x$ and square it:

$x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2$

Then bring the constant $c$ back in:

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

Watch out: $(x+a)^2 = x^2 + 2ax + a^2$, not $x^2 + a^2$. The middle term $2ax$ is crucial.

Completing the square (a ≠ 1)

When the coefficient of $x^2$ is not 1, factor it out first:

$ax^2 + bx + c = a\left(x^2 + \frac{b}{a}x\right) + c$

Now complete the square inside the bracket, then multiply through by $a$:

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

This step is where most errors happen. Take your time factoring out $a$, and always expand to check.

The discriminant

Consider the quadratic equation $ax^2 + bx + c = 0$. Using the quadratic formula:

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

The expression under the square root, $\Delta = b^2 - 4ac$, is called the discriminant. It controls how many real roots the equation has, because we can only take the square root of a non-negative number.

This is not a random formula to memorise — it falls straight out of completing the square. If you complete the square on $ax^2 + bx + c = 0$ and rearrange, you arrive at $\sqrt{b^2 - 4ac}$ naturally.

Discriminant summary

Language note: At AS level we say “no real roots” rather than “no roots”. The equation still has solutions — you’ll meet them if you study Further Maths — but they are not real numbers.

Worked Examples

Example 1: Completing the square (a = 1)

Write $x^2 + 8x + 3$ in the form $(x + p)^2 + q$.

Half the coefficient of $x$ is $\frac{8}{2} = 4$. So:

$x^2 + 8x + 3 = (x + 4)^2 - 16 + 3 = (x + 4)^2 - 13$

The vertex of the parabola is at $(-4,\, -13)$.

Example 2: Solving by completing the square

Solve $x^2 + 6x + 1 = 0$, giving your answer in surd form.

Complete the square:

$x^2 + 6x + 1 = (x + 3)^2 - 9 + 1 = (x + 3)^2 - 8$

Set equal to zero:

$(x + 3)^2 = 8$

$x + 3 = \pm\sqrt{8} = \pm 2\sqrt{2}$

$x = -3 + 2\sqrt{2} \quad \text{or} \quad x = -3 - 2\sqrt{2}$

Don’t forget the ±. The square root step always gives two values. Dropping the negative root is one of the most common errors in this topic.

Example 3: Completing the square (a ≠ 1)

Write $2x^2 + 12x + 7$ in the form $a(x + p)^2 + q$.

Factor out 2 from the first two terms:

$2x^2 + 12x + 7 = 2(x^2 + 6x) + 7$

Complete the square inside the bracket:

$= 2\left[(x + 3)^2 - 9\right] + 7$

Expand and simplify:

$= 2(x + 3)^2 - 18 + 7 = 2(x + 3)^2 - 11$

Check: $2(x+3)^2 - 11 = 2(x^2 + 6x + 9) - 11 = 2x^2 + 12x + 18 - 11 = 2x^2 + 12x + 7$ ✓

Example 4: Using the discriminant

Show that $x^2 + 3x + 5 = 0$ has no real roots.

Here $a = 1$, $b = 3$, $c = 5$.

$b^2 - 4ac = 9 - 20 = -11$

Since $b^2 - 4ac < 0$, the equation has no real roots. $\square$

The parabola $y = x^2 + 3x + 5$ sits entirely above the $x$-axis.

Example 5: Discriminant with an unknown (exam-style)

The equation $kx^2 + 6x + k = 0$ has two distinct real roots. Find the range of values of $k$, where $k \neq 0$.

For two distinct real roots we need $b^2 - 4ac > 0$.

Here $a = k$, $b = 6$, $c = k$:

$36 - 4(k)(k) > 0$

$36 - 4k^2 > 0$

$4k^2 < 36$

$k^2 < 9$

$-3 < k < 3$

Since $k \neq 0$ (otherwise it wouldn’t be a quadratic), the final answer is:

$-3 < k < 3, \quad k \neq 0$

Common Patterns

These standard results come up repeatedly at AS level:

Key link to graphs: If the completed square form is $(x + p)^2 + q$, the line of symmetry is $x = -p$ and the minimum value of the quadratic is $q$ (when $a > 0$). If $a < 0$, the parabola is upside-down and $q$ is the maximum.