Completing the square, discriminant, solving quadratic equations and inequalities.
Answer at least 3 of 3 correctly to complete this section.
At GCSE you factorised quadratics like x2+5x+6=(x+2)(x+3). But try factorising x2+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.
These two skills — completing the square and using the discriminant — appear in almost every AS exam paper. Master them here and you’ll use them throughout the course.
We want to rewrite x2+bx+c in the form (x+p)2+q, where the vertex of the parabola sits at (−p,q).
Start with the x2+bx part. We take half the coefficient of x and square it:
x2+bx=(x+2b)2−(2b)2
Then bring the constant c back in:
x2+bx+c=(x+2b)2−(2b)2+c
Watch out: (x+a)2=x2+2ax+a2, not x2+a2. The middle term 2ax is crucial.
Write x2+8x+3 in the form (x+p)2+q.
Half the coefficient of x is 28=4. So:
x2+8x+3=(x+4)2−16+3=(x+4)2−13
The vertex of the parabola is at (−4,−13).
Solve x2+6x+1=0, giving your answer in surd form.
Complete the square:
x2+6x+1=(x+3)2−9+1=(x+3)2−8
Set equal to zero:
(x+3)2=8
x+3=±8=±22
x=−3+22orx=−3−22
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.
Answer at least 3 of 4 correctly to complete this section.
When the coefficient of x2 is not 1, factor it out first:
ax2+bx+c=a(x2+abx)+c
Now complete the square inside the bracket, then multiply through by a:
=a(x+2ab)2−4ab2+c
This step is where most errors happen. Take your time factoring out a, and always expand to check.
Write 2x2+12x+7 in the form a(x+p)2+q.
Factor out 2 from the first two terms:
2x2+12x+7=2(x2+6x)+7
Complete the square inside the bracket:
=2[(x+3)2−9]+7
Expand and simplify:
=2(x+3)2−18+7=2(x+3)2−11
Check: 2(x+3)2−11=2(x2+6x+9)−11=2x2+12x+18−11=2x2+12x+7 ✓
Write 3x2+12x+5 in the form a(x+p)2+q.
Hint: What is the first step? (Factor out 3 from the x2 and x terms before completing the square inside the bracket.)
Factor out 3: 3(x2+4x)+5
Complete the square inside: 3[(x+2)2−4]+5
Expand: 3(x+2)2−12+5=3(x+2)2−7
Check: 3(x+2)2−7=3(x2+4x+4)−7=3x2+12x+12−7=3x2+12x+5 ✓
Enter a=1, b=8, c=3 into the visualiser below — these are the values from Worked Example 1. Step through and verify that you get the same completed square form.
Enter values for $a$, $b$, $c$ and step through the process.
Step through these four expressions. Watch how the completed-square form changes.
Half of 2 is 1.
Answer at least 3 of 4 correctly to complete this section.
Consider the quadratic equation ax2+bx+c=0. Using the quadratic formula:
x=2a−b±b2−4ac
The expression under the square root, Δ=b2−4ac, 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 ax2+bx+c=0 and rearrange, you arrive at b2−4ac naturally.
| Condition | Meaning | What the graph does |
|---|---|---|
| b2−4ac>0 | Two distinct real roots | Crosses the x-axis twice |
| b2−4ac=0 | One repeated real root | Touches the x-axis (tangent) |
| b2−4ac<0 | No real roots | Does not meet the x-axis |
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.
Adjust the sliders below. Can you find values where the parabola just touches the x-axis? What is the discriminant value when that happens?
Adjust the sliders to see how the discriminant affects the roots.
Show that x2+3x+5=0 has no real roots.
Here a=1, b=3, c=5.
b2−4ac=9−20=−11
Since b2−4ac<0, the equation has no real roots. □
The parabola y=x2+3x+5 sits entirely above the x-axis.
Answer at least 3 of 4 correctly to complete this section.
These examples combine completing the square with the discriminant — exactly as they appear on AS exam papers.
The equation kx2+6x+k=0 has two distinct real roots. Find the range of values of k, where k=0.
For two distinct real roots we need b2−4ac>0.
Here a=k, b=6, c=k:
36−4(k)(k)>0
36−4k2>0
4k2<36
k2<9
−3<k<3
Since k=0 (otherwise it wouldn’t be a quadratic), the final answer is:
−3<k<3,k=0
These standard results come up repeatedly at AS level:
| Expression | Completed square form | Vertex |
|---|---|---|
| x2+bx+c | (x+2b)2+c−4b2 | (−2b,c−4b2) |
| x2−6x+10 | (x−3)2+1 | (3,1) |
| −(x2−4x)+7 | −(x−2)2+11 | (2,11) — maximum |
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.
Answer at least 4 of 5 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.
Simplifying surds, laws of indices, rationalising denominators.
Solving linear and quadratic inequalities, representing solutions on a number line and using set notation.
Algebraic division, factor theorem, remainder theorem, factorising cubics.
Translations, stretches, and reflections of graphs. Effect of transformations on equations.
Equation of a circle, finding tangents and normals, circle properties and intersection with lines.
First principles, differentiating x^n, gradient functions, rates of change.