What is a quadratic equation in the context of IB MYP Mathematics?

In the IB MYP Mathematics curriculum, a quadratic equation is a type of polynomial equation of the form ax^2 + bx + c = 0, where 'a', 'b', and 'c' are constants and 'a' is not equal to zero. Quadratic equations are fundamental in Algebra and are used to model various real-world scenarios.

How do you solve a quadratic equation using the quadratic formula?

To solve a quadratic equation using the quadratic formula, you apply the formula x = (-b ± √(b^2 - 4ac)) / (2a). This formula provides the solutions for x in the equation ax^2 + bx + c = 0. It is essential to calculate the discriminant (b^2 - 4ac) first to determine the nature of the roots.

What is the significance of the discriminant in a quadratic equation?

The discriminant in a quadratic equation, given by b^2 - 4ac, indicates the nature of the roots. If the discriminant is positive, the equation has two distinct real roots. If it is zero, there is one real root (a repeated root). If the discriminant is negative, the equation has two complex roots. Understanding the discriminant helps in predicting the type of solutions without solving the equation.

Can you explain how to factor a quadratic equation?

Factoring a quadratic equation involves expressing it as a product of two binomials. For example, the equation x^2 + 5x + 6 can be factored into (x + 2)(x + 3) = 0. To factor, you need to find two numbers that multiply to the constant term 'c' and add up to the coefficient 'b'. This method is effective when the quadratic is easily factorable.

What are some real-world applications of quadratic equations?

Quadratic equations are used in various real-world applications, such as calculating projectile motion, optimizing areas and volumes, and modeling economic problems. In the IB MYP Mathematics curriculum, students learn to apply quadratic equations to solve practical problems, enhancing their understanding of how algebra is used in everyday life.

Why is "Trying to factorise $x^2 + 5x = 14$ as if it equals 0." a common mistake in Quadratic Equations?

Why it happens: Skipping the 'all on one side' step. How to avoid it: Always move everything to one side first: x^2 + 5x - 14 = 0, then factorise.

Why is "Forgetting the minus sign in $-b$ in the quadratic formula." a common mistake in Quadratic Equations?

Why it happens: Mental arithmetic shortcut. How to avoid it: Write down each value of a, b, c FIRST, with their signs. Then substitute carefully into the formula.

Why is "Missing the negative root." a common mistake in Quadratic Equations?

Why it happens: Stopping after the first solution. How to avoid it: A quadratic typically has TWO roots. The in the formula gives both. Always state BOTH (unless the discriminant is 0).

IB MYP Mathematics (MYP - M)

Quadratic Equations

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Short Notes - Quadratic Equations

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Detailed Study Notes

Detailed notes on Algebra for IB MYP Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Quadratic Equations — IB MYP Mathematics (Extended): factorisation, formula, and the discriminant

A quadratic equation has the form $ax^2 + bx + c = 0$ . This MYP Mathematics Extended note covers three solution methods (factorisation, completing the square, formula), the discriminant, and applications.

At a glance

A QUADRATIC equation: $ax^2 + bx + c = 0$ with $a \ne 0$ .
Three solution methods: FACTORISATION, COMPLETING THE SQUARE, QUADRATIC FORMULA.
Quadratic formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
DISCRIMINANT $\Delta = b^2 - 4ac$ tells you the number of real roots.
$\Delta > 0$ : 2 real roots; $\Delta = 0$ : 1 repeated root; $\Delta < 0$ : no real roots.

What you’ll learn

Mapped to the IB MYP Mathematics subject guide (2026 onwards).

MYP Mathematics A — Solve quadratics by factorisation.
MYP Mathematics A — Apply the quadratic formula.
MYP Mathematics A — Use the discriminant to classify roots.
MYP Mathematics D — Model real situations with quadratics.

Solving by factorisation

Easiest method when factorisation works.

If you can factorise the quadratic into a product of two brackets, you can solve it by the zero-product rule: if $AB = 0$ , then $A = 0$ or $B = 0$ .

Worked example. Solve $x^2 - 5x + 6 = 0$ .

Factorise: $x^2 - 5x + 6 = (x - 2)(x - 3)$ .
Zero-product: $x - 2 = 0$ or $x - 3 = 0$ .
So $x = 2$ or $x = 3$ .

Worked example. Solve $2x^2 + 7x + 3 = 0$ .

Factorise (AC method): $2x^2 + 7x + 3 = (2x + 1)(x + 3)$ .
$2x + 1 = 0$ → $x = -\tfrac{1}{2}$ . $x + 3 = 0$ → $x = -3$ .
So $x = -\tfrac{1}{2}$ or $x = -3$ .

Worked example (difference of squares). Solve $x^2 - 9 = 0$ .

$x^2 - 9 = (x - 3)(x + 3) = 0$ .
$x = 3$ or $x = -3$ .

(Alternatively: $x^2 = 9 \Rightarrow x = \pm 3$ .)

Always:

Move everything to one side (= 0).
Factorise.
Set each factor to 0.
Solve for $x$ .

Move all terms to one side.
Factorise the quadratic.
Zero-product rule: each bracket = 0.

The quadratic formula

Works EVERY time.

When factorisation is hard or impossible, use the quadratic formula:

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

It comes from completing the square on the general form $ax^2 + bx + c = 0$ .

Worked example. Solve $2x^2 + 5x - 3 = 0$ using the formula.

$a = 2$ , $b = 5$ , $c = -3$ .
Discriminant: $b^2 - 4ac = 25 - 4(2)(-3) = 25 + 24 = 49$ .
$x = \dfrac{-5 \pm \sqrt{49}}{4} = \dfrac{-5 \pm 7}{4}$ .
$x = \dfrac{2}{4} = 0.5$ or $x = \dfrac{-12}{4} = -3$ .

Worked example. Solve $x^2 - 4x + 1 = 0$ .

$a = 1$ , $b = -4$ , $c = 1$ .
$b^2 - 4ac = 16 - 4 = 12$ .
$x = \dfrac{4 \pm \sqrt{12}}{2} = \dfrac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$ .

This is a case where factorisation would fail — the roots are irrational. Always have the formula in your toolkit.

Strategy: try factorisation first (faster); use the formula if it doesn't factorise nicely.

Formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
Works for ANY quadratic.
Try factorisation first; formula as a backup.

The discriminant

Tells you how many roots without solving.

The expression under the square root in the formula is called the discriminant:

$\Delta = b^2 - 4ac.$

It controls the number of real roots:

$\Delta$	Number of real roots	Geometric meaning
$> 0$	2 distinct real roots	parabola crosses x-axis twice
$= 0$	1 repeated root (or 'double root')	parabola touches x-axis once (tangent)
$< 0$	0 real roots	parabola never crosses x-axis

The discriminant $\Delta = b^2 - 4ac$ controls how many times the parabola $y = ax^2 + bx + c$ crosses the x-axis.

Worked example. How many roots does $x^2 + 6x + 9 = 0$ have?

$\Delta = 36 - 36 = 0$ → ONE repeated root.
Factorise: $(x + 3)^2 = 0 \Rightarrow x = -3$ (double root).

Worked example. Does $x^2 + 2x + 5 = 0$ have real roots?

$\Delta = 4 - 20 = -16 < 0$ → NO real roots.

Many Extended-level questions ask you to use the discriminant to find unknown coefficients.

$\Delta = b^2 - 4ac$ .
Positive $\to 2$ roots; zero $\to 1$ repeated; negative $\to 0$ real roots.
Tells you the number of x-axis crossings without solving.

Quick recap

Standard form: $ax^2 + bx + c = 0$ .
Methods: factorisation, formula, completing the square.
Formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
Discriminant $\Delta = b^2 - 4ac$ : classifies number of real roots.

Memorise this

Verbatim phrases and definitions MYP criterion-A markschemes credit.

Quadratic: $ax^2 + bx + c = 0$
Formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$\Delta = b^2 - 4ac$
$\Delta > 0$ : 2 roots; $=0$ : 1; $<0$ : 0

How it’s examined

Criterion A: solve by factorisation or formula. Criterion C: clear step-by-step solving. Criterion D: model a real situation with a quadratic (projectile, area).

Sources: IB MYP Mathematics subject guide (IBO, official). Last reviewed 2026-05-25.

Take this whole topic with you

Step-by-step worked examples — Quadratic Equations

Step-by-step solutions to past-paper-style questions on quadratic equations, written exactly the way a tutor would explain them at the board.

1Solve by factorisation
Building confidence• factorise
Question
Solve $x^2 + 8x + 15 = 0$ .
Step-by-step solution
1. Step 1
  Find two numbers multiplying to 15 and adding to 8: $3 + 5 = 8$ , $3 \times 5 = 15$ .
2. Step 2
  Factorise: $(x + 3)(x + 5) = 0$ .
3. Step 3
  Zero-product: $x = -3$ or $x = -5$ .
Answer
$x = -3$ or $x = -5$ .
2Use the quadratic formula
Building confidence• formula
Question
Solve $3x^2 - 7x + 2 = 0$ .
Step-by-step solution
1. Step 1
  $a = 3$ , $b = -7$ , $c = 2$ .
2. Step 2
  $\Delta = 49 - 24 = 25$ , so $\sqrt{\Delta} = 5$ .
3. Step 3
  $x = \dfrac{7 \pm 5}{6}$ . So $x = 2$ or $x = \tfrac{1}{3}$ .
Answer
$x = 2$ or $x = \tfrac{1}{3}$ .
3Find $k$ using the discriminant
Stretch• discriminant
Question
Find the values of $k$ for which $x^2 + kx + 9 = 0$ has exactly one repeated root.
Step-by-step solution
1. Step 1
  One repeated root means $\Delta = 0$ .
2. Step 2
  $\Delta = k^2 - 36 = 0$ . So $k^2 = 36$ , $k = \pm 6$ .
Answer
$k = 6$ or $k = -6$ .
4Real context — projectile
Stretch• application
Question
A ball thrown upwards has height $h = -5t^2 + 20t + 1$ metres after $t$ seconds. When does it hit the ground?
Step-by-step solution
1. Step 1
  Hits ground when $h = 0$ : $-5t^2 + 20t + 1 = 0$ .
2. Step 2
  Multiply by $-1$ (no flip): $5t^2 - 20t - 1 = 0$ .
3. Step 3
  Formula: $a = 5$ , $b = -20$ , $c = -1$ . $\Delta = 400 + 20 = 420$ . $\sqrt{420} \approx 20.5$ .
4. Step 4
  $t = \dfrac{20 \pm 20.5}{10}$ . Two roots: $\approx 4.05$ s or $-0.05$ s.
5. Step 5
  Reject negative time. Ball hits ground at $t \approx 4.05\,\text{s}$ .
Answer
$t \approx 4.05$ s.

Key Definitions and Keywords — Quadratic Equations

Definitions to memorise and the exact keywords mark schemes credit for quadratic equations answers — sharpened from recent examiner reports for the 2026 IB MYP Mathematics (Extended) sitting.

Quadratic equation
Examiner keyword
An equation of the form $ax^2 + bx + c = 0$ with $a \ne 0$ .
Root (of an equation)
Examiner keyword
A value of $x$ that satisfies the equation. For a quadratic, also called a solution or zero.
Discriminant $\Delta$
Examiner keyword
$\Delta = b^2 - 4ac$ . Determines the number of real roots.
Completing the square
Rewriting $ax^2 + bx + c$ as $a(x - h)^2 + k$ — useful for finding the vertex of a parabola.

Common Mistakes and Misconceptions — Quadratic Equations

The traps other students keep falling into on quadratic equations questions — taken from recent IB MYP Mathematics (Extended) examiner reports and mark schemes — and how to avoid them.

✕Trying to factorise $x^2 + 5x = 14$ as if it equals 0.
Why it happens
Skipping the 'all on one side' step.
How to avoid it
Always move everything to one side first: $x^2 + 5x - 14 = 0$ , then factorise.
✕Forgetting the minus sign in $-b$ in the quadratic formula.
Why it happens
Mental arithmetic shortcut.
How to avoid it
Write down each value of $a, b, c$ FIRST, with their signs. Then substitute carefully into the formula.
✕Missing the negative root.
Why it happens
Stopping after the first solution.
How to avoid it
A quadratic typically has TWO roots. The $\pm$ in the formula gives both. Always state BOTH (unless the discriminant is 0).

Quadratic Equations — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

When factorisation is hard or impossible, use the quadratic formula:

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

It comes from completing the square on the general form $ax^2 + bx + c = 0$ .

Worked example. Solve $2x^2 + 5x - 3 = 0$ using the formula.

$a = 2$ , $b = 5$ , $c = -3$ .
Discriminant: $b^2 - 4ac = 25 - 4(2)(-3) = 25 + 24 = 49$ .
$x = \dfrac{-5 \pm \sqrt{49}}{4} = \dfrac{-5 \pm 7}{4}$ .
$x = \dfrac{2}{4} = 0.5$ or $x = \dfrac{-12}{4} = -3$ .

Worked example. Solve $x^2 - 4x + 1 = 0$ .

$a = 1$ , $b = -4$ , $c = 1$ .
$b^2 - 4ac = 16 - 4 = 12$ .
$x = \dfrac{4 \pm \sqrt{12}}{2} = \dfrac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$ .

This is a case where factorisation would fail — the roots are irrational. Always have the formula in your toolkit.

Strategy: try factorisation first (faster); use the formula if it doesn't factorise nicely.

\Delta

Number of real roots

Geometric meaning

> 0

2 distinct real roots

parabola crosses x-axis twice

= 0

1 repeated root (or 'double root')

parabola touches x-axis once (tangent)

< 0

0 real roots

parabola never crosses x-axis

Quadratic Equations

Short Notes - Quadratic Equations

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Solve by factorisation

2Use the quadratic formula

3Find kkk using the discriminant

4Real context — projectile

Quadratic equation

Root (of an equation)

Discriminant Δ\DeltaΔ

Completing the square

✕Trying to factorise x2+5x=14x^2 + 5x = 14x2+5x=14 as if it equals 0.

✕Forgetting the minus sign in −b-b−b in the quadratic formula.

✕Missing the negative root.

Quadratic Equations — frequently asked questions

Quadratic Equations

Short Notes - Quadratic Equations

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Solve by factorisation

2Use the quadratic formula

3Find kkk using the discriminant

4Real context — projectile

Quadratic equation

Root (of an equation)

Discriminant Δ\DeltaΔ

Completing the square

✕Trying to factorise x2+5x=14x^2 + 5x = 14x2+5x=14 as if it equals 0.

✕Forgetting the minus sign in −b-b−b in the quadratic formula.

✕Missing the negative root.

Quadratic Equations — frequently asked questions

3Find $k$ using the discriminant

Discriminant $\Delta$

✕Trying to factorise $x^2 + 5x = 14$ as if it equals 0.

✕Forgetting the minus sign in $-b$ in the quadratic formula.

3Find $k$ using the discriminant

Discriminant $\Delta$

✕Trying to factorise $x^2 + 5x = 14$ as if it equals 0.

✕Forgetting the minus sign in $-b$ in the quadratic formula.