Introduction to Solving Quadratic Equations

A quadratic equation is any equation that can be written in the standard form:

$a x^{2} + b x + c = 0$

where $a \neq = 0$ , and $a$ , $b$ , $c$ are real constants.

Solving a quadratic means finding the values of $x$ (called roots or solutions) that satisfy the equation. There are three main methods you need to know for IB Maths AA SL:

Factoring , fast and elegant, but only works cleanly for certain quadratics
Completing the square , a powerful algebraic technique that also reveals the vertex form
The quadratic formula , always works, derived directly from completing the square

Exam Tip

On IB exams, your GDC can solve quadratics numerically. However, you must also be able to solve them by hand, especially in Paper 1 (no calculator). Knowing all three methods gives you flexibility.

Method 1: Factoring

Factoring relies on the fact that if a quadratic has roots $r_{1}$ and $r_{2}$ , then it can be written as:

$a x^{2} + b x + c = a (x - r_{1}) (x - r_{2})$

Expanding this gives:

$a (x^{2} - (r_{1} + r_{2}) x + r_{1} r_{2})$

Comparing coefficients with $a x^{2} + b x + c$ , we see:

$b = - a (r_{1} + r_{2})$ , so $r_{1} + r_{2} = - \frac{b}{a}$
$c = a \cdot r_{1} r_{2}$ , so $r_{1} r_{2} = \frac{c}{a}$

The key trick: when $a = 1$ , you need two numbers that multiply to $c$ and add to $b$ .

Example

Solve $x^{2} - 5 x + 6 = 0$ by factoring.

We need two numbers that multiply to $6$ and add to $- 5$ .

Those numbers are $- 2$ and $- 3$ , since $(- 2) (- 3) = 6$ and $(- 2) + (- 3) = - 5$ .

So: $x^{2} - 5 x + 6 = (x - 2) (x - 3) = 0$

Therefore $x = 2$ or $x = 3$ .

Example

Solve $2 x^{2} + 7 x + 3 = 0$ by factoring.

Here $a = 2$ , so we look for two numbers that multiply to $a c = 2 \times 3 = 6$ and add to $b = 7$ .

Those numbers are $1$ and $6$ . Split the middle term:

$2 x^{2} + x + 6 x + 3 = x (2 x + 1) + 3 (2 x + 1) = (x + 3) (2 x + 1) = 0$

Therefore $x = - 3$ or $x = - \frac{1}{2}$ .

Warning

Not every quadratic factors neatly over the integers. If you cannot find integer factors quickly, switch to the quadratic formula rather than wasting exam time.

Free preview

9 more sections in this topic

Introduction to Solving Quadratic Equations

A quadratic equation is any equation that can be written in the standard form:

$a x^{2} + b x + c = 0$

where $a \neq = 0$ , and $a$ , $b$ , $c$ are real constants.

Solving a quadratic means finding the values of $x$ (called roots or solutions) that satisfy the equation. There are three main methods you need to know for IB Maths AA SL:

Factoring , fast and elegant, but only works cleanly for certain quadratics
Completing the square , a powerful algebraic technique that also reveals the vertex form
The quadratic formula , always works, derived directly from completing the square

Exam Tip

On IB exams, your GDC can solve quadratics numerically. However, you must also be able to solve them by hand, especially in Paper 1 (no calculator). Knowing all three methods gives you flexibility.

Method 1: Factoring

Factoring relies on the fact that if a quadratic has roots $r_{1}$ and $r_{2}$ , then it can be written as:

$a x^{2} + b x + c = a (x - r_{1}) (x - r_{2})$

Expanding this gives:

$a (x^{2} - (r_{1} + r_{2}) x + r_{1} r_{2})$

Comparing coefficients with $a x^{2} + b x + c$ , we see:

$b = - a (r_{1} + r_{2})$ , so $r_{1} + r_{2} = - \frac{b}{a}$
$c = a \cdot r_{1} r_{2}$ , so $r_{1} r_{2} = \frac{c}{a}$

The key trick: when $a = 1$ , you need two numbers that multiply to $c$ and add to $b$ .

Example

Solve $x^{2} - 5 x + 6 = 0$ by factoring.

We need two numbers that multiply to $6$ and add to $- 5$ .

Those numbers are $- 2$ and $- 3$ , since $(- 2) (- 3) = 6$ and $(- 2) + (- 3) = - 5$ .

So: $x^{2} - 5 x + 6 = (x - 2) (x - 3) = 0$

Therefore $x = 2$ or $x = 3$ .

Example

Solve $2 x^{2} + 7 x + 3 = 0$ by factoring.

Here $a = 2$ , so we look for two numbers that multiply to $a c = 2 \times 3 = 6$ and add to $b = 7$ .

Those numbers are $1$ and $6$ . Split the middle term:

$2 x^{2} + x + 6 x + 3 = x (2 x + 1) + 3 (2 x + 1) = (x + 3) (2 x + 1) = 0$

Therefore $x = - 3$ or $x = - \frac{1}{2}$ .

Warning

Not every quadratic factors neatly over the integers. If you cannot find integer factors quickly, switch to the quadratic formula rather than wasting exam time.

Free preview

SL 2.7—Solutions of quadratic equations and inequalities, discriminant and nature of roots

Introduction to Solving Quadratic Equations

Method 1: Factoring

9 more sections in this topic

SL 2.7—Solutions of quadratic equations and inequalities, discriminant and nature of roots

Introduction to Solving Quadratic Equations

Method 1: Factoring

9 more sections in this topic