Learn · Mathematics · Class 10

Quadratic Equations

Three solution methods, the discriminant, and choosing the right approach.

11 parts

Scroll to begin ↓

📘 DefinitionThe standard formPart 1 of 11

🧩 The partsThree methodsPart 2 of 11

The roots are where the curve meets the x-axis.

✏️ Worked exampleSolving by factorisingPart 4 of 11

🔍 In detailThe discriminantPart 5 of 11

✏️ Worked exampleCompleting the squarePart 6 of 11

✏️ Worked exampleA worded problemPart 7 of 11

🌍 In the real worldWhere quadratics appearPart 8 of 11

⚠️ Common mistakesWhere marks are lostPart 9 of 11

✅ SummaryWhat to rememberPart 10 of 11

✏️ Worked exampleWatch one worked throughPart 11 of 11

1📘 DefinitionPart 1 / 11

The standard form

A quadratic equation has a squared term as its highest power.

ax² + bx + c = 0, a ≠ 0

If a were zero the x² term vanishes and it becomes linear, which is why the condition is stated.

A quadratic has at most two solutions, called roots. Graphically these are where the parabola crosses the x-axis.

Always rearrange to = 0 first. Every method assumes it.

2🧩 The partsPart 2 / 11

Three methods

Factorising: fastest when it works. Try it first.
Completing the square: always works, and gives the turning point as a by-product.
The formula: always works. Use it when factorising fails.

x = [−b ± √(b² − 4ac)] / 2a

How to choose: attempt factorising for about thirty seconds. If no integer pair works, go straight to the formula. Persisting with factorising on a non-factorising quadratic wastes exam time.

3🔍 In detailPart 3 / 11

Why factorising solves it

The method rests on one property of zero, worth stating explicitly.

If a product equals zero, at least one factor must be zero.

(x − 3)(x + 5) = 0
→ x − 3 = 0 or x + 5 = 0
→ x = 3 or x = −5

This only works for zero. If (x−3)(x+5) = 8, you cannot conclude anything about the factors. There are infinitely many pairs multiplying to 8.

That is precisely why the equation must be rearranged to = 0 before factorising. Skipping that step is a guaranteed wrong answer.

4✏️ Worked examplePart 4 / 11

Solving by factorising

Question. Solve 2x² − 5x − 3 = 0.

Step 1: a × c. 2 × (−3) = −6.

Step 2: two numbers multiplying to −6, adding to −5.

−6 and +1 (−6 × 1 = −6, −6 + 1 = −5) ✅

Step 3: split the middle term.

2x² − 6x + x − 3 = 0

Step 4: group and factorise each pair.

2x(x − 3) + 1(x − 3) = 0
(x − 3)(2x + 1) = 0

Step 5: set each factor to zero.

x = 3 or x = −1/2

Check both in the original. Giving only one root loses half the marks, a very common slip.

5🔍 In detailPart 5 / 11

The discriminant

The expression under the root sign is the discriminant.

Δ = b² − 4ac

It tells you the nature of the roots without solving.

Δ > 0: two distinct real roots. The curve crosses the axis twice.
Δ = 0: one repeated root. The curve touches the axis.
Δ < 0: no real roots. The curve never reaches the axis.

Where this is examined: "for what values of k does this equation have equal roots?" Set Δ = 0 and solve for k. These questions are common and are answered entirely by the discriminant.

6✏️ Worked examplePart 6 / 11

Completing the square

Question. Solve x² + 6x − 7 = 0 by completing the square.

Step 1: halve the x coefficient and square it. Half of 6 is 3, squared is 9.

Step 2: write as a perfect square, subtracting what you added.

(x + 3)² − 9 − 7 = 0
(x + 3)² − 16 = 0

Step 3: rearrange and take the square root.

(x + 3)² = 16
x + 3 = ±4

The ± is essential, omitting it loses one root, and it is the most common error in this method.

x = 1 or x = −7

The by-product: the form (x + 3)² − 16 tells you the turning point is at (−3, −16) immediately. Neither factorising nor the formula gives you that.

7✏️ Worked examplePart 7 / 11

A worded problem

Question. A rectangular garden is 3 m longer than it is wide, with an area of 88 m². Find its dimensions.

Step 1: define the variable. Let the width be x. Length = x + 3.

Step 2: form the equation.

x(x + 3) = 88
x² + 3x − 88 = 0

Step 3: factorise. Two numbers multiplying to −88, adding to 3: 11 and −8.

(x + 11)(x − 8) = 0
x = −11 or x = 8

Step 4: reject the impossible root. A width cannot be negative, so x = 8.

Answer: width 8 m, length 11 m.

Step 4 is where the marks are. A quadratic gives two roots; a physical problem often admits only one. Stating why you rejected the other earns the mark, silently discarding it does not.

8🌍 In the real worldPart 8 / 11

Where quadratics appear

Projectile motion. The path of anything thrown is a parabola, so height against time is a quadratic. Solving it finds when the object lands.

Area and optimisation. Maximising an enclosed area for a fixed perimeter is a quadratic problem.

Economics. Profit against price is often quadratic, which is why there is an optimum price rather than "as high as possible".

Design. Parabolic shapes focus signals to a point, which is why satellite dishes and headlight reflectors have that curve.

The general pattern: quadratics appear wherever one quantity depends on the square of another, which is far more often than it first appears.

9⚠️ Common mistakesPart 9 / 11

Where marks are lost

Not rearranging to = 0 before factorising.
Giving only one root.
Omitting the ± when taking a square root.
Sign errors in the formula: especially −b when b is already negative.
Discarding a root silently in a worded problem.
Persisting with factorising when the roots are not rational.

10✅ SummaryPart 10 / 11

What to remember

Rearrange to = 0; a product is zero only if a factor is.
Try factorising briefly, then use the formula.
Δ = b² − 4ac gives the nature of the roots.
Two roots, and justify rejecting one in a real problem.

11✏️ Worked examplePart 11 / 11

Watch one worked through

Here is one done slowly, with the reason for every move. Then one for you.

Watch one first

Solve 2x² - 7x + 3 = 0 using the formula.

Identify a, b and c from the standard form.
a = 2, b = −7, c = 3
Write them down with their signs. Losing the minus on b is the single commonest error in this whole topic.
Work out the discriminant first, on its own.
b² − 4ac = (−7)² − 4(2)(3) = 49 − 24 = 25
Doing it separately keeps the arithmetic clean, and its sign tells you how many roots to expect before you go further.
Substitute into the formula.
x = (7 ± √25) ÷ (2 × 2) = (7 ± 5) ÷ 4
Note −b is +7 because b was negative. The whole numerator is over 2a, so keep it bracketed.
Take both signs in turn.
x = 12/4 = 3 x = 2/4 = 0.5
A quadratic has two roots. Stopping at one loses half the marks even when the one you found is right.

Answer: x = 3 and x = 0.5

Now you try

For x² - 5x + 6 = 0, what is the discriminant b² - 4ac?

Show the answer

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