Factorising Quadratics 2 - Intermediate - Numbas at mathcentre.ac.uk

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9 years, 7 months ago

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Katie Lester 9 years, 7 months ago

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Katie Lester 9 years, 7 months ago

Created this as a copy of Factorising Quadratics 1 - Basic.

Name	Status	Author	Last Modified
Factorising Quadratics 1 - Basic	Ready to use	Katie Lester	01/06/2016 09:47
Factorising Quadratics 2 - Intermediate	Ready to use	Katie Lester	01/06/2016 09:47
Factorising Quadratics 3 - Challenging	Ready to use	Katie Lester	01/06/2016 09:47
Factorising Quadratics 4 - Solving by Factorising or Using Formula	Ready to use	Katie Lester	01/06/2016 09:47
Factoring 2 Quadratics 2 of form: $ax^2+bx+c$	draft	Clare Lundon	05/10/2016 22:45
Factorising Quadratics 1: $x^2+bx+c$ 3 questions READY	draft	Clare Lundon	05/10/2016 22:46
steve's copy of Factoring Quadratics 2 $ax^2+bx+c$ 4 questions READY	draft	steve kilgallon	18/07/2016 16:05
Lucy's copy of Factoring Quadratics 2 $ax^2+bx+c$ 4 questions READY	draft	Lucy Hadfield	03/08/2016 10:50
Francis 's copy of Solving Quadratic Equations using the Quadratic Formula 4 Questions READY	draft	Francis Duah	12/08/2016 17:56
Gareth's copy of Lucy's copy of Factoring Quadratics 2 $ax^2+bx+c$ 4 questions READY	draft	Gareth Woods	26/08/2016 11:21
Kjell's copy of Francis 's copy of Solving Quadratic Equations using the Quadratic Formula 4 Questions READY	draft	Kjell Hansen	05/09/2016 11:01
Kjell's copy of Kjell's copy of Francis 's copy of Solving Quadratic Equations using the Quadratic Formula 4 Questions READY	draft	Kjell Hansen	05/09/2016 11:02
Solving Quadratic Equations using the Quadratic Formula and a Calculator	draft	Francis Duah	14/09/2016 16:55
Francis 's copy of Solving Quadratic Equations using the Quadratic Formula 4 Questions READY	draft	Francis Duah	14/09/2016 16:50
Factoring 1 Quadratic of form: $ax^2+bx+c$	draft	Clare Lundon	13/10/2016 10:06
Solving Quadratic Equations using the Quadratic Formula	draft	Hannah Bartholomew	18/09/2017 11:52
Ugur's copy of Factorising Quadratics 2 - Intermediate	Ready to use	Ugur Efem	01/11/2021 12:31
Ugur's copy of Factorising Quadratics 1 - Basic	Ready to use	Ugur Efem	04/11/2021 15:33

There are 19 other versions that do you not have access to.

Question statement

Simplify the following quadratics into two linear factors in the form $(ax+c)(bx+d)$ .

Quadratics are defined as having 'repeated roots' if the both roots in the solution are the same. This can be written as $(x+a)(x+a)$ , but can be simplified to $(x+a)^2$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	1
b	integer	1
c	integer	4
d	integer	4
f1	integer	1
f2	integer	8
f3	integer	16

			Name	Type	Generated Value

a2	integer	2
b2	integer	2
c2	integer	4
d2	integer	4
f12	integer	4
f22	integer	16
f32	integer	16

			Name	Type	Generated Value

a3	integer	3
b3	integer	3
c3	integer	-2
d3	integer	6
f13	integer	9
f23	integer	12
f33	integer	-12

			Name	Type	Generated Value

a4	integer	3
b4	integer	2
c4	integer	0
d4	integer	3
f14	integer	6
f24	integer	9
f34	integer	0

			Name	Type	Generated Value

Automatically regenerate variables?

Name

Data type

Value

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1

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `integer`

→ Used by:

Advice
"Part a)" → "Unnamed gap" - Correct answer

Parts

Part a) Gap-fill
1. Gap Mathematical expression
  Add
Part b) Gap-fill
1. Gap Mathematical expression
  Add
Part c) Gap-fill
1. Gap Mathematical expression
  Add
Part d) Gap-fill
1. Gap Mathematical expression
  Add
1. Step Information only

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

$\simplify{{f1}x^2+{f2}*x+{f3}}$

$=$ Unnamed gap

Advice

Repeated roots questions, using the equation from Part a:

$\simplify{{f1}x^2+{f2}*x+{f3}}=0$

These are solved in the exact same way as the previous set of quadratics, but the 'roots' (solutions) are the same.

This type of equation is described as having 'repeated roots'.

The graphed functions would only touch the $x$ -axis once either at their peak or trough.

Looking at the above equation, you may be able to instantly see that the brackets must both include the quantity $\var{c}$ .

If not, don't worry, just go through the same process of trial and error as taught previously until you reach your solution of

$\simplify{({a}x+{c})({b}x+{d})}$ .

Then, noticing this can be further simplified, write the final simplification as:

$\simplify{({a}x+{c})^2}$ .

Note: If the $x^2$ term has a coefficient and simplifies to, for example, $(2x+4)(x+4)$ , this is not defined as having repeated roots, since the solutions are:

$2x+4=0$ so $x=\frac{-4}{2}=-2$

and

$x+4=0$ so $x=-4$

so are in fact different.

Zero constants equations, using Part d:

$\simplify{{f14}x^2+{f24}*x+{f34}}=0$

Some equations will not have a constant at the end, meaning one of the constants is zero.

In this case, if there is still an $x$ term, you know one of the constants must still be non-zero.

You must therefore, take out a factor of a number $\times x$ .

Here, the factor could be $\simplify{{a4}x}$ or $\simplify{{b4}x}$ .

It doesn't matter which you choose, but in this example we will use $\simplify{{a4}x}$ .

In order to pull out this factor, the rest of the equation must be divided by $\simplify{{a4}x}$ :

$\frac{\simplify{{f14}x^2+{f24}*x+{f34}}}{\simplify{{a4}*x}}=\simplify{{f14}x/{a4}+{f24}/{a4}}$ .

Therefore, the solution is:

$\simplify{{a4}x*({f14}x/{a4}+{f24}/{a4})}$ .

Give a worked solution to the whole question.

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Factorising Quadratics 2 - Intermediate

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Katie Lester 9 years, 7 months ago

Katie Lester 9 years, 7 months ago

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Generated value: `integer`

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Factorising Quadratics 2 - Intermediate

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Katie Lester 9 years, 7 months ago

Katie Lester 9 years, 7 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`