GCSE · Maths · AQA · Spec 8300 · Foundation

Deduce roots of quadratics algebraically

Draw y = x² − 2x − 8 and it crosses the x-axis twice. The surprise? You can pin down both crossing points without drawing anything at all.

Where does y = x² − 2x − 8 cross the x-axis?

x: 1. y: -9

You're starting at the lowest point, (1, −9). Drag left and right along the curve and stop each time y reads exactly 0.

Line of symmetry, x = 1: Fold the graph along this line and the two halves match exactly. It sits halfway between the crossings (−2 and 4 average to 1), and the turning point (1, −9) is right on it. Check it: x = −1 and x = 3 are both 2 steps from the line, and both give y = −5.

Predict it with algebra

Turn x² − 2x − 8 into two brackets

The graph told us the answers. Now let's predict them. The corners are done: x × x = x², and the two missing numbers multiply to give −8. Find two numbers that multiply to −8 AND add to −2. Put the positive one on the side and the negative one along the top, fill the two middle cells (together they must make −2x), then write the brackets.

Grid: each cell is its row times its column
×	x
x	x²
		−8

So x² − 2x − 8 =

Type x² as x^2 if you cannot type ². Spaces do not matter.

Why zero matters

What if it doesn't equal zero?

Jas has to solve (x − 1)(x − 4) = 10.

What should Jas do? Pick the idea closest to what you'd do.

Set each bracket equal to 10. So x − 1 = 10 or x − 4 = 10, giving x = 11 or x = 14.The answers are still x = 1 and x = 4. The brackets tell you the roots whatever is on the right.10 = 2 × 5, so x − 1 = 2 and x − 4 = 5.Expand, then move the 10 across so one side is 0. Then factorise again.

How sure are you?

I'm sureI think soI'm guessing

Set each bracket equal to 10. So x − 1 = 10 or x − 4 = 10, giving x = 11 or x = 14.

This is a common misconception

It feels like the same move as before, but test it. Put x = 11 in: (11 − 1)(11 − 4) = 10 × 7 = 70, not 10. The bracket rule never said 'each bracket equals the right-hand side'. It only works with 0, because a product of 0 forces one factor to be 0. A product of 10 forces nothing.

The answers are still x = 1 and x = 4. The brackets tell you the roots whatever is on the right.

This is a common misconception

Try x = 1: (1 − 1)(1 − 4) = 0 × (−3) = 0. That solves (x − 1)(x − 4) = 0, not = 10. The brackets tell you where the curve y = (x − 1)(x − 4) crosses the x-axis. This equation asks where that curve reaches a height of 10, which is somewhere else.

10 = 2 × 5, so x − 1 = 2 and x − 4 = 5.

This is a common misconception

That gives x = 3 from one bracket and x = 9 from the other, but x can't be two different numbers at once. And 10 can be made in endless ways: 1 × 10, 4 × 2.5, −2 × −5… There's no way to know which pair the brackets are. Zero is special: the only way to multiply two numbers and get 0 is for one of them to be 0.

Expand, then move the 10 across so one side is 0. Then factorise again.

This matches the evidence

Exactly. (x − 1)(x − 4) = 10 becomes x² − 5x + 4 = 10, then x² − 5x − 6 = 0. That factorises to (x − 6)(x + 1) = 0, so x = 6 or x = −1. Check: (6 − 1)(6 − 4) = 5 × 2 = 10 ✓ and (−1 − 1)(−1 − 4) = (−2)(−5) = 10 ✓. Notice it's 'or': x is 6 or −1, and only one bracket is 0 at a time.

Zero first. Only a product of 0 tells you something about each bracket, so rearrange to '= 0' before you factorise and split.

Spot the slip

Where does this answer go wrong?

Solve x² + 2x − 15 = 0, then find the line of symmetry of the curve y = x² + 2x − 15.

A student's working — which line goes wrong?

x² + 2x − 15 = 0
5 × (−3) = −15 and 5 + (−3) = 2, so (x + 5)(x − 3) = 0
x = 5 or x = −3
Line of symmetry is halfway: (5 + (−3)) ÷ 2 = 1
So the line of symmetry is x = 1

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

Predict the crossing points with algebra. Then watch the graph agree.

What you need to know

A root of a quadratic is a value of x that makes it equal 0. On the graph, the roots are where the curve meets the x-axis.
To find roots algebraically: get the quadratic equal to 0, factorise it, then set each bracket equal to 0.
To factorise x² + bx + c, find two numbers that multiply to c and add to b.
A quadratic graph is symmetrical. Its line of symmetry is halfway between the roots, and the turning point sits on that line.

The big picture

The roots of a quadratic are the values of x that make it equal zero, so on a graph they are where the curve meets the x-axis (y = 0). To find them without a graph, rearrange so the quadratic equals 0, factorise it, then ask what makes each bracket zero. This works because two numbers can only multiply to 0 if one of them is 0. Watch the signs: x + 2 = 0 gives x = −2. The graph is symmetrical, so its line of symmetry sits halfway between the roots, with the turning point on it.

Key points

1Roots = x-intercepts = where y = 0.

2Zero first: only a product of 0 forces one bracket to be 0.

3For a bracket like (x + 2), flip the sign: the root is x = −2.

4Line of symmetry: x = (first root + second root) ÷ 2.

.we-_R_18slubsnpfiv5ubrb_ [data-we-step]{display:flex !important}.we-_R_18slubsnpfiv5ubrb_ [data-we-solution]{display:block !important}.we-_R_18slubsnpfiv5ubrb_ [data-we-reveal]{display:none !important}

Worked example

Problem

The curve y = x² − 8x + 12. Find where it crosses the x-axis, and the coordinates of its turning point.

⚠ Watch out

Reading the root straight off the bracket. (x + 3)(x − 7) = 0 does NOT give x = 3 or x = −7. Ask what makes each bracket zero: x + 3 = 0 when x = −3, and x − 7 = 0 when x = 7.

🧠

Memory hook

Zero it, split it, flip it: make it = 0, factorise into brackets, then ask what makes each bracket zero. For (x + 2), that's x = −2.

✓

Check yourself

Find the roots of x² + 3x − 10 = 0. Then say where the line of symmetry of y = x² + 3x − 10 is. Check your roots by substituting them back in.

Flashcards

(6)

What is a root of a quadratic?

A value of x that makes the quadratic equal 0. On the graph it's where the curve meets the x-axis (where y = 0).

Before you factorise to find roots, what must the equation look like?

Quadratic = 0. If the other side isn't 0, rearrange first.

Why does (x − a)(x − b) = 0 mean x = a or x = b?

Two numbers can only multiply to make 0 if one of them is 0. So x − a = 0 or x − b = 0.

Which root does the bracket (x + 6) give?

x = −6. Ask what makes the bracket zero: x + 6 = 0 when x = −6.

A quadratic has roots p and q. Where is its line of symmetry?

Halfway between them: x = (p + q) ÷ 2. The turning point lies on this line.

To factorise x² + bx + c, what pair of numbers do you look for?

Two numbers that multiply to c and add to b.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

Learning with Lightbulb is opening soon

You can use this lesson now. Join the waitlist and we'll let you know when the full Lightbulb experience is ready.

Keep me posted

Where does y = x² − 2x − 8 cross the x-axis?

Turn x² − 2x − 8 into two brackets

What if it doesn't equal zero?

Where does this answer go wrong?

More AQA GCSE Maths topics