Solve quadratic inequalities

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Christian Lawson-Perfect	said	Ready to use	7 years, 9 months ago
Bradley Bush	said	Needs to be tested	7 years, 9 months ago
Vicky Hall	said	Has some problems	7 years, 9 months ago

History

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Elliott Fletcher 7 years, 8 months ago

Published this.

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Christian Lawson-Perfect 7 years, 9 months ago

Gave some feedback: Ready to use

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Christian Lawson-Perfect 7 years, 9 months ago

Saved a checkpoint:

I've fixed a couple of typos, but otherwise this is OK!

Restore this checkpoint

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Bradley Bush commented 7 years, 9 months ago

Thank you for the feedback, I can see your logic, so I've changed it to your way of asking.

The inequality signs have been swapped to make sure the student is paying attention to their direction because of it's significance to the question.

Both graphs struggled to fit on the same figure so I have added another graph at the end with an brief bit of text to justify that both x ranges are the same.

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Bradley Bush 7 years, 9 months ago

Gave some feedback: Needs to be tested

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Christian Lawson-Perfect commented 7 years, 9 months ago

(I realise I contradicted Vicky's suggestion. Not sure if what you've done is exactly what she imagined, but it doesn't read clearly to me!)

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Christian Lawson-Perfect 7 years, 9 months ago

Gave some feedback: Has some problems

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Christian Lawson-Perfect commented 7 years, 9 months ago

I've changed "sketch the quadratic" to "sketch each quadratic".

The prompt doesn't actually say what should go in the first gap for each part. I know the statement says factorise first, but you should be absolutely clear what you want.

I'd have:

"Factorise $f(x)$ :

$x^2+ax+b =$ [[gap]]

Hence, find the range of values for $x$ such that $x^2+ax+b \gt 0$ ."

You need a conjunction between the two inequalities at the end of parts a and b: " $x \gt ??$ OR $x \lt ??$ " in part a, and $x \gt ??$ AND $x \lt ??$ " for part b. If you want, you can make that a dropdown.

The inequality signs for part b are the other way round to part a. Why?

Part c doesn't make sense - you can't rearrange an inequality to get an equality. I'd ask them to rearrange to get something $\gt 0$ - this also makes sure they don't give the negation of what you're expecting - and then say "by factorising or otherwise, give the range of values for which $ax + b \gt x^2$ . Keeping track of the inequality makes it easier to work out which way round the upper and lower bounds should go.

In the advice, I'd give examples of some values of $x$ that satisfy the inequality - in part a, a big negative value and a big positive value, and in part b a vlaue between the roots. For part c, I think it would be very helpful to show the graphs of $x^2$ and $ax+b$ on top of the graph of $x^2-ax-b$ to show that the inequality holds for the same values in both arrangements - i.e., $x^2 \gt ax+b$ exactly when $x^2-ax-b \gt 0$ .

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Bradley Bush 7 years, 9 months ago

Gave some feedback: Needs to be tested

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Bradley Bush 7 years, 9 months ago

Gave some feedback: Has some problems

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Bradley Bush 7 years, 9 months ago

Gave some feedback: Needs to be tested

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Bradley Bush commented 7 years, 9 months ago

Thank you for the feedback, those typos were careless and hopefully they are all sorted now, I have also reworded the question for you.

I hate to disagree but I think that it would be useful for students who are struggling in particular to have the factorise step left in the question to give the student a mid way point. But if there is more logic behind removing the gap, I am happy to take it out.

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Vicky Hall 7 years, 9 months ago

Gave some feedback: Has some problems

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Vicky Hall commented 7 years, 9 months ago

Reword the questions here to say something like: find the range of values of $x$ for which ${function}>0$ . You have told the student to factorise in the statement so you don't need to say it again. I would also remove the gap asking them to give the factorised form.

Have a check through the advice as I noticed a couple of typos and a rogue part d).

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Bradley Bush 7 years, 9 months ago

Gave some feedback: Needs to be tested

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Bradley Bush commented 7 years, 9 months ago

How do I allow either or answers for the gap in part a?

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Bradley Bush 7 years, 9 months ago

Created this as a copy of Solving linear inequalities.

Name	Status	Author	Last Modified
Solving linear inequalities	Ready to use	Bradley Bush	20/11/2019 14:42
Solve quadratic inequalities	Ready to use	Bradley Bush	20/11/2019 14:42
Solving linear inequalities	Should not be used	Adrian Jannetta	31/08/2023 17:00
Ida's copy of Solve quadratic inequalities	draft	Ida Landgärds	13/04/2018 09:16
Ida's copy of Solving linear inequalities	draft	Ida Landgärds	13/04/2018 09:16
Solving linear inequalities [L5 Randomised]	Needs to be tested	Matthew James Sykes	25/07/2018 10:51
cormac's copy of Solving linear inequalities	draft	cormac breen	24/10/2018 13:44
cormac's copy of Solve quadratic inequalities	draft	cormac breen	24/10/2018 13:46
Solving linear inequalities	draft	Xiaodan Leng	11/07/2019 01:43

There are 66 other versions that do you not have access to.

Question statement

Solve the following quadratic inequalities by firstly factorising $f(x)$ and then solving for $x$ when $f(x)=0$. It may be helpful to sketch each quadratic.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	2
b	integer	2
c	integer	3
d	integer	2
g	integer	4
f	integer	5

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Name

Data type

Value

xxxxxxxxxx
 
random(1..5)

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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)" - prompt
"Part a)" → "Gap 0." - Correct answer
"Part a)" → "Gap 2." - Correct answer

Gap-fill

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

Find the range of values for $x$ such that $\simplify{f(x)=x^2+({a}-{b})x-{a}{b}}>0$.

Factorise $f(x)$:

$\simplify{f(x)=x^2+({a}-{b})x-{a}{b}}=$ Gap 0. $=0$.

Hence,

$x>$ Gap 1.

Gap 3.

$x<$ Gap 2.

Advice

a)

This example is best illustrated by the graph of $\simplify{f(x)=x^2+({a}-{b})x-{a}{b}}$ below. By finding the roots of the equation we can find the $x$ coordinates where the line crosses the $x$ axis and then we can use a sketch or visualise the graph to work out the set of values for $x$ where $f(x)>0$.\[\\[0.1em]\]

{geogebra_applet('Hk5dTptY',[["a",a],["b",b]])}

\[
\begin{align}
\simplify{f(x)=x^2+({a}-{b})x-{a}{b}}&>0\\
\simplify{(x+{a})(x-{b})}&=0\text{.}
\end{align}
\]

Therefore

\[ \simplify{x<{-a}}\text{ or }\simplify{x>{b}}\text{.}\]

b)

We use the same method in this example but this time we use our graph to visualise where $g(x)<0$.\[\\[0.1em]\]

{geogebra_applet('PUFStTNa',[["c",c],["d",d]])}

\[
\begin{align}
\simplify{f(x)=x^2+({c}-{d})x-{c}{d}}&<0\\
\simplify{(x+{c})(x-{d})}&=0\text{.}
\end{align}
\]

Therefore

\[ \simplify{x>{-c}}\text{ and }\simplify{x<{d}}\text{.}\]

c)

The notable difference with solving this equation is the requirement to rearrange the inequality before factorisation.\[\\[0.1em]\]

{geogebra_applet('CZsSCdH6',[["e",g],["f",f]])}

\[
\begin{align}
\simplify{({g}+{f})x-{g}{f}}&>x^2\\
\simplify{x^2-({g}+{f})x+{g}{f}}&<0\\
\simplify{(x-{g})(x-{f})}&=0\text{.}
\end{align}
\]

Therefore

\[ \simplify{x>{g}}\text{ and }\simplify{x<{f}}\text{.}\]

Alternatively, we can plot the graph of $x^2$ against $\simplify{({g}+{f})x-{g}{f}}$ to visualise the same result.

{geogebra_applet('Ez697SNE',[["g",g],["f",f]])}

From this graph we can see that the values of $x$ where $\simplify{({g}+{f})x-{g}{f}}>x^2$ are the same as the values of $x$ where $\simplify{x^2-({g}+{f})x+{g}{f}}<0$.

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This question is used in the following exams:

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Week 1 Diagnostic test by Anne-Sophie Kaloghiros in MA1630 - Fundamentals of Mathematics 2024-2025.

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Elliott Fletcher 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Bradley Bush commented 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Christian Lawson-Perfect commented 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Christian Lawson-Perfect commented 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Bradley Bush commented 7 years, 9 months ago

Vicky Hall 7 years, 9 months ago

Vicky Hall commented 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Bradley Bush commented 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Gap-fill

a)

b)

c)

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Solve quadratic inequalities

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Elliott Fletcher 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Bradley Bush commented 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Christian Lawson-Perfect commented 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Christian Lawson-Perfect commented 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Bradley Bush commented 7 years, 9 months ago

Vicky Hall 7 years, 9 months ago

Vicky Hall commented 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Bradley Bush commented 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

a)

b)

c)

Generated value: `integer`