Simon's copy of Factorising Quadratic Equations with $x^2$ Coefficients of 1

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Simon Thomas 6 years ago

Created this as a copy of Factorising Quadratic Equations with

Name	Status	Author	Last Modified
Factorising Quadratic Equations with $x^2$ Coefficients of 1	Ready to use	Hannah Aldous	20/11/2019 14:39
Solving Quadratic Equations with $x^2$ Coefficients of 1 by Factorising	Should not be used	Hannah Aldous	20/11/2019 14:42
Katherine's copy of Factorising Quadratic Equations with $x^2$ Coefficients of 1	draft	Katherine Tomlinson	02/08/2017 12:44
Peter's copy of Factorising Quadratic Equations with $x^2$ Coefficients of 1	draft	Peter Knowles	12/09/2017 12:29
MATH6058 Factorising Quadratic Equations with $x^2$ Coefficients of 1	draft	Catherine Palmer	27/01/2020 17:30
Ida's copy of Factorising Quadratic Equations with $x^2$ Coefficients of 1	draft	Ida Landgärds	13/04/2018 09:15
Ida's copy of Solving Quadratic Equations with $x^2$ Coefficients of 1 by Factorising	Doesn't work	Ida Landgärds	19/04/2018 12:00
løs og faktorisere andregradslikninga	draft	Ida Landgärds	02/12/2020 15:42
spørsmål 6	draft	Ida Landgärds	02/12/2020 15:42
Quadratic Formula	draft	Blathnaid Sheridan	19/09/2024 11:51
Blathnaid's copy of spørsmål 6	draft	Blathnaid Sheridan	02/12/2020 15:42
Blathnaid's copy of spørsmål 6	draft	Blathnaid Sheridan	02/12/2020 15:42
Ed's copy of Factorising Quadratic Equations with $x^2$ Coefficients of 1	draft	Ed Southwood	08/08/2022 10:18
Factorising Quadratic Expressions	draft	Ed Southwood	08/08/2022 10:18
Simon's copy of Factorising Quadratic Equations with $x^2$ Coefficients of 1	draft	Simon Thomas	07/03/2019 09:52
what the heck?	draft	John Wick	30/05/2019 11:33
Factorizing Quadratic Equations with $x^2$ Coefficients of 1	draft	Xiaodan Leng	10/07/2019 21:45
Factorizing Quadratic Equations with $x^2$ Coefficients of 1	draft	Xiaodan Leng	10/07/2019 21:45
Musa's copy of 3 Factorising Quadratic Equations with $x^2$ Coefficients of 1	draft	Musa Mammadov	27/06/2023 01:26
Ugur's copy of Factorising Quadratic Equations with $x^2$ Coefficients of 1	Ready to use	Ugur Efem	08/11/2021 16:44
Factorising quadratics EXAM	Ready to use	Yvonne Wancke	17/04/2023 11:42
Factorising quadratics S1MT2	Ready to use	Yvonne Wancke	22/03/2023 11:35
AD1 Factorising a Quadratic (a=1)	Ready to use	Andrew Neate	09/04/2024 15:13

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Question statement

Factorise the following quadratic expressions.

Give any introductory information the student needs.

			Name	Type	Generated Value

v1	integer	4
v2	integer	3
v3	integer	-3
v4	integer	10
v5	integer	8
v6	integer	-8

			Name	Type	Generated Value

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Generated value: `integer`

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

Parts

Part a) Gap-fill
1. Gap Mathematical expression
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Part b) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Number entry
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3. Gap Gap 2. Number entry
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Part c) Gap-fill
1. Gap Mathematical expression
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Gap-fill

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

$\simplify{x^2+{v1+v2}x+{v1*v2}}$

Unnamed gap

Advice

Quadratic expressions of the form

$x^2+bx+c$ can be factorised to create an equation of the form

$(x+m)(x+n)\text{.}$

Note also that

$(x+m)(x+n)=x^2+(m+n)x+(m \times n)\text{.}$

Thus to factorise an expression of the form $x^2+bx+c$ , we need to find two numbers which add together to make $b$ , and multiply together to make $c$ .

(a)

$\simplify{x^2+{v1+v2}x+{v1*v2}}$

We need to find two values that add together to make $\var{v1+v2}$ and multiply together to make $\var{v1*v2}$ .

$\begin{align} \var{v1} \times \var{v2}&=\var{v1*v2}\\ \var{v1}+\var{v2}&=\var{v1+v2}\\ \end{align}$

So the factorised form of the expression is

$\simplify{(x+{v1})(x+{v2})}\text{.}$

(b)

We need to find two numbers that multiply to give $\var{v3*v4}$ and add together to give $\var{v3+v4}$ .

$\begin{align} \var{v3} \times \var{v4}&=\var{v3*v4}\\ \var{v3}+\var{v4}&=\var{v3+v4}\\ \end{align}$

So the factorised form of the expressionis

$\simplify{(x+{v3})(x+{v4})}\text{.}$

Hence to solve $\simplify{x^2+{v3+v4}x+{v3*v4}}=0$ we have:

$\simplify{(x+{v3})(x+{v4})}=0$

So either $\simplify{(x+{v3})}=0$ or $\simplify{(x+{v4})}=0$

which gives solutions of $x=\var{-v3}$ or $x=\var{-v4}$

(c)

When factorising the quadratic expression

$\simplify{x^2+{v5*v6}}$

we need to find two values that add together to make $0$ and multiply together to make $\var{v5*v6}$ .

$\begin{align} \var{v5} \times \var{v6}& = \var{v5*v6}\\ \simplify[]{ {v5} + {v6}} &= 0 \\ \end{align}$

So the factorised form of the expression is

$\simplify{(x+{v5})(x+{v6})}\text{.}$

Note this is a common form of quadratic called a "Difference of Two Squares". In general $x^2-a^2=(x-a)(x+a)$

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This question is used in the following exam:

Quadratic expressions and equations by Simon Thomas in Algebra.

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Simon's copy of Factorising Quadratic Equations with $x^2$ Coefficients of 1

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Simon Thomas 6 years ago

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Generated value: `integer`

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Simon's copy of Factorising Quadratic Equations with x2x^2 Coefficients of 1

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

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Simon Thomas 6 years ago

Error in variable testing condition

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Generated value: integer

Parts

Gap-fill

(a)

(b)

(c)

Simon's copy of Factorising Quadratic Equations with $x^2$ Coefficients of 1

Generated value: `integer`