# Equations » Quadratic equations (x2 = number / sum-product-method / the quadratic formula)

## Recognise:

Quadratic equations can be divided into three categories.
Every category has its own way of solving the equation.

 Category: Examples: Reducible to x2 = number x2 = 25  or  x2 – 7 = 2 Reducible to binomial = 0 5x2 – x = 0 or 3x2 + 5x = x2 – 7x Reducible to trinomial = 0 5x2 – 4x + 3 = 0 or 3x2 = 2x2 + 4x + 12

Factorising formulas (incl. sum-product-method)
Completing the square
Calculate the vertex
Shape of the parabola
Calculate the points of intersection with the axes
More examples (mixed)

## x2 = number

Use the balance method or the card method to solve these equations.
Reduce to x2 on the left-hand side and then find the square root.
REMEMBER: MOST OFTEN THERE ARE TWO SOLUTIONS!

Example 1
 x2 = 25 x = or x = – x = 5 or x = –5

Example 2
 x2 + 7 = 2 x2 = –5 doesn't exist:
no solutions

Example 3
 –2x2 + 1 = –97 –2x2 = –98 x2 = 49 x = or x = – x = 7 or x = –7
Example 4
 (x + 3)2 + 5 = 86 (x + 3)2 = 81 x + 3 = – or x + 3 = x + 3 = –9 or x + 3 = 9 x = –12 or x = 6

## Binomial = 0

Reduce to zero and factorise the binomial (single brackets). After factorising, either the left-hand factor or the right-hand factor must be zero.

Example 1
 5x2 – x = 0 x(5x – 1) = 0
 x = 0 or 5x – 1 = 0 x = 0 or 5x = 1 x = 0 or x = 15
Example 2
 3x2 + 5x = x2 – 7x 2x2 + 12x = 0 2x(x + 6) = 0
 2x = 0 or x + 6 = 0 x = 0 or x = –6

## Trinomial = 0

Reduce to zero and then factorise the trinomial (double brackets). When factorising is not possible, use the quadratic formula / abc-formula or completing the square.

Example 1
 2x2 + 4x + 12 = 3x2 –x2 + 4x + 12 = 0 x2 – 4x – 12 = 0 (x + 2)(x – 6) = 0
 x + 2 = 0 or x – 6 = 0 x = –2 or x = 6
Example 2
5x2 – 4x – 3 = 0
a = 5,  b = –4,  c = –3
D = (–4)2 – 4 · 5 · –3 = 76
 x = –(–4) + 2 × 5 or x = –(–4) – 2 × 5 x ≈ 1.27 or x ≈ –0.47

## Factorising

### Single brackets (binomial)

Look for the greatest common divider in both terms.
That is what you write in front of the brackets.
Your factorised formula is always of the form y = …(… + …)  or  y = …(… – …).
Example: In both terms of y = 6x2 + 15x you have a factor 3x
After all, the formula can be written as y = 2 × 3 × x × x + 3 × 5 × x.
This formula can therefore be factorised and written as y = 3x(2x + 5).

### Sum-product-method (trinomials)

A sum is the answer of an addition and a product is the answer of a multiplication.
The sum-product-method can only be used with a trinomial that is
written as x2 + bx + c = 0 (so a = 1).
Underneath this trinomial you write double brackets like this: (x … …)(x … …) = 0
Behind the x on the open spaces you have to fill in the two numbers that are added b and multiplied c. Write a + in front of positive numbers.

Example 1
 x2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 x = –3 or x = –2

Example 2
 x2 – 6x + 9 = 0 (x – 3)(x – 3) = 0
x = 3

Example 3
15x2 + 15x – 6 = 0
get a = 1 by multiplying every term by 5
x2 + x – 30 = 0
(x – 5)(x + 6) = 0
x = 5 or x = –6
Example 4
5x2 + 25x – 70 = 0
get a = 1 by dividing every term by 5
x2 + 5x – 14 = 0
(x – 2)(x + 7) = 0
x = 2 or x = –7

## The quadratic formula / abc-formula

This formula can be used to solve every quadratic equation.
However, only use it on trinomials that cannot be factorised, as it costs a lot of work.

The equation has to be written as ax2 + bx + c = 0.
Find the numbers for a, b and c. Watch out with negative numbers!

With these numbers you first calculate the Discriminant, the formula is:
D = b2 – 4ac
When D is negative, you have no solutions.
When D is zero, you have one solution.
When D is positive, you have two solutions.

When D is not negative, use the following two formulas to calculate the values of x.
x = b + 2a or x = b – 2a

Watch out! On your calculator this might look like this:
(–b + ) / (2 × a)

Do you want to know how this formula is derived?
Look at derivation of the quadratic formula.

Example 2 at 'Trinomial = 0' above and example 6 all the way at the bottom of the page are examples of the quadratic formula.

## Completing the square

You can also solve a quadratic equation by completing the square. However, it is not necessary to learn this. After all, you can solve any quadtratic equation with the quadratic formula / abc-formula. Look at completing the square for the theory and a number of examples on how you can solve a quadratic equation by completing the square.

## Calculate the vertex

Use xvertex = b2a.
To calculate yvertexyou have to fill in xvertexinto the formula.

## Shape of the parabola

Is completely determined by a.
When a = positive you have an upward-opening parabola
When a = negative you have a downward-opening parabola
The bigger the difference between a and zero, the narrower the parabola is.

## Calculate the points of intersection with the axes

For the intersection with the y-axis, calculate y for x = 0.
For the intersections with the x-axis calculate x for y = 0.

## Quadratic equations with a parameter

When you have an equation like 2x2 + 5x + p = 0  or  –x2 + 2x  + 1,5 = –4x + p, check quadratic equations with a parameter.

## More examples (mixed)

Example 1
 x2 – 5 = 31 x2 = 36 x = or x = – x = 6 or x = –6

Example 2
 (5 – 2x)2 = 81 5 – 2x = or 5 – 2x = – 5 – 2x = 9 or 5 – 2x = –9 –2x = 4 or –2x = –14 x = –2 or x = 7

Example 3
 2x2 = 8x 2x2 – 8x = 0 2x(x – 4) = 0 2x = 0 or x – 4 = 0 x = 0 or x = 4

Example 4
 x(x + 2) = 3 x2 + 2x = 3 x2 + 2x – 3 = 0 (x + 3)(x – 1) = 0 x + 3 = 0 or x – 1 = 0 x = –3 or x = 1

Example 5
 (3x – 1)2 = (x – 7)2 3x – 1 = x – 7 or 3x – 1 = –(x – 7) 2x = –6 or 3x – 1 = –x + 7 x = –3 or 4x = 8 x = –3 or x = 2

Example 6
 x2 + 7x + 9.5 = 19x – 2x2 3x2 – 12x + 9.5 = 0
a = 3,  b = –12,  c = 9.5
D = (–12)2 – 4 × 3 × 9.5 = 30
 x = –(–12) + 2 × 3 or x = –(–12) – 2 × 3 x ≈ 2.91 or x ≈ 1.09

To top