Rafa?'s copy of Solving a quadratic equation

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Rafa? Porzezinski

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Rafa? Porzezinski 7 years, 2 months ago

Created this as a copy of Solving a quadratic equation.

Name	Status	Author	Last Modified
Solving equations	draft	Bill Foster	06/08/2020 08:41
Solving a quadratic equation	draft	Lois Rollings	29/07/2016 13:42
Francis 's copy of Solving a quadratic equation	draft	Francis Duah	12/08/2016 17:55
Rafa?'s copy of Solving a quadratic equation	draft	Rafa? Porzezinski	11/01/2018 14:14

There are 27 other versions that do you not have access to.

Question statement

Znajdź pierwiastki podanego równania kwadratowego

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	4
c	integer	7
b	integer	4
d	integer	8
f	integer	16
s3	integer	1
s2	integer	-1
s1	integer	-1
n4	number	4
n2	integer	-4
disc	number	16
rdis	string	The roots are real and distinc
n1	integer	60
c1	integer	7
n3	integer	32
rep	string
n5	integer	16
d1	integer	7

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..5)

JME function reference

Description

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

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

d
disc
f
n1
n2
n3
n5

This variable doesn't seem to be used anywhere.

Parts

Gap-fill
1. Gap Gap 0. Mathematical expression
  Add
2. Gap Gap 1. Mathematical expression
  Add

Gap-fill

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

$\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0$

Podja wartości pierwiastków jako ułamki lub liczby całkowite, nie dziesiętne

Wpisz pierwiastek z niższą wartością jako pierwszy, jeżeli równanie posiada jeden podwójny pierwiastek wpisz go w obu miejscach.

Pierwiuastek z niższą wartością $x=\;$ Gap 0..

Pierwiastek z wyższa wartością $x=\;$ Gap 1.

Advice

There are different ways to solve a quadratic equation.

If you can see how to factorise the quadratic expression then this should be the quickest method.

For this example we have

$\simplify{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d} = ({a} * x + { -c}) * ({b} * x + { -d})}$

So either

$\simplify{ ({a} * x + { -c}) =0} \ \text{or} \ \ \simplify{ ({b} * x + { -d})}=0$

So the roots are:

$x= \simplify{{n1-n4}/{2*a*b}} \ \ \text{and} \ \ x= \simplify{{n1+n4}/{2*a*b}}$

If you cannot see a factorisation then you can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Remember that there are three possible types of solution, depending upon the value of the discriminant $\Delta=b^2-4ac$ .

1. $\Delta \gt 0$ . The roots are real and distinct

2. $\Delta=0$ . The roots are real and equal. Their common value is $\displaystyle -\frac{b}{2a}$

3. $\Delta \lt 0$ . There are no real roots. (The roots are complex)

For this question the discriminant is $\Delta = \simplify[std]{{-n1}^2-4*{a*b*c*d}}=\var{disc}$ . {rdis}

So the {rep} roots are:

$x = \frac{\var{n1} - \sqrt{\var{disc}}}{\var{n3}} = \frac{\var{n1} - \var{n4} }{\var{n3}} = \simplify{{n1 - n4}/ {n3}} \ \ \text{and} \ \ x = \frac{\var{n1} + \sqrt{\var{disc}}}{\var{n3}} = \frac{\var{n1} + \var{n4} }{\var{n3}} = \simplify{{n1 + n4}/ {n3}}$

You could also use the method of completing the square.

First we write:

$\begin{eqnarray} \simplify{{a*b}x^2+{-n1}x+{c*d}}&=&\var{n5}\left(\simplify{x^2+({-n1}/{a*b})x+ {c*d}/{a*b}}\right)\\ &=&\var{n5}\left(\left(\simplify{x+({-n1}/{2*a*b})}\right)^2+ \simplify{{c*d}/{a*b}-({-n1}/({2*a*b}))^2}\right)\\ &=&\var{n5}\left(\left(\simplify{x+({-n1}/{2*a*b})}\right)^2 -\simplify{ {n2^2}/{4*(a*b)^2}}\right) \end{eqnarray}$
So we need to solve:

$\begin{eqnarray} \left(\simplify{x+({-n1}/{2*a*b})}\right)^2&\phantom{{}}& -\simplify{ {n2^2}/{4*(a*b)^2}}=0\Rightarrow\\ \left(\simplify{x+({-n1}/{2*a*b})}\right)^2&\phantom{{}}&=\simplify{ {n2^2}/{4*(a*b)^2}=({abs(n2)}/{2*a*b})^2} \end{eqnarray}$
and get the two {rep} solutions:

$\begin{eqnarray} \simplify{x+({-n1}/{2*a*b})}&=&\simplify{-{abs(n2)}/{2*a*b}} \Rightarrow &x& = \simplify{({-abs(n2)+n1}/{2*a*b})}\\ \simplify{x+({-n1}/{2*a*b})}&=&\simplify{({abs(n2)}/{2*a*b})} \Rightarrow &x& = \simplify{({n1+abs(n2)}/{2*a*b})} \end{eqnarray}$

See also these Mathcentre leaflets on solving quadratic equations.

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This question is used in the following exam:

Zadanie 1 by Rafa? Porzezinski in Rafa?'s workspace.

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Rafa?'s copy of Solving a quadratic equation

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Rafa? Porzezinski 7 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

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Gap-fill

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Rafa?'s copy of Solving a quadratic equation

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Rafa? Porzezinski 7 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`