Solving equations

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Bill Foster

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Bill Foster 55 years, 2 months ago

Created this.

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

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Question statement

Find the roots of the following quadratic equation.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	5
c	integer	7
b	integer	4
d	integer	3
f	integer	20
s3	integer	1
s2	integer	-1
s1	integer	1
n5	integer	20
disc	number	169
rep	string
rdis	string	The roots exist and are distin
n1	integer	43
c1	integer	7
n3	integer	40
n4	number	13
n2	integer	13
d1	integer	3

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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:

d
disc
f
n1
n2
n3
n5

Advice
"Unnamed part" - prompt
"Unnamed part" → "Gap 0." - Correct answer
"Unnamed part" → "Gap 1." - Correct answer

Parts

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

Solve for $x$ :

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

$x=\;$ Gap 0.. The greatest root is

$x=\;$ Gap 1.

You can get more information on solving a quadratic by clicking on Show steps. You will lose 1 mark if you do so.

Enter the least root first. If the roots are equal, enter the root in both input boxes.

Enter the roots as fractions or integers, not as decimals.

Advice

Direct Factorisation

If you can spot a direct factorisation then this is the quickest way to do this question.

For this example we have the factorisation

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

Hence we find the roots:

$\begin{eqnarray} x&=& \simplify{{n1-n4}/{2*a*b}}\\ x&=& \simplify{{n1+n4}/{2*a*b}} \end{eqnarray}$

Other Methods.

There are several methods of finding the roots – here are the main methods.

Finding the roots of a quadratic using the standard formula.

We can use the following formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

The two roots are

$x = \frac{-b +\sqrt{b^2-4ac}}{2a}\mbox{ and } x = \frac{-b -\sqrt{b^2-4ac}}{2a}$
there are three main types of solutions 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 root are complex and form a complex conjugate pair.

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

{rdis}.

So the {rep} roots are:

$\begin{eqnarray} x = \frac{\var{n1} - \sqrt{\var{disc}}}{\var{n3}} &=& \frac{\var{n1} - \var{n4} }{\var{n3}} &=& \simplify{{n1 - n4}/ {n3}}\\ x = \frac{\var{n1} + \sqrt{\var{disc}}}{\var{n3}} &=& \frac{\var{n1} + \var{n4} }{\var{n3}} &=& \simplify{{n1 + n4}/ {n3}} \end{eqnarray}$

Completing the square.

First we complete the square for the quadratic expression $\simplify{{a*b}x^2+{-n1}x+{c*d}}$

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

$\simplify{{a*b}x^2+{-n1}x+{c*d}}=0$ we have to solve:

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

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

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Information only

This question is used in the following exams:

mathcentre: Factorising quadratics by Bill Foster in Bill's workspace.
Andy Pomfret's Test by Andy Pomfret in Andy's workspace.
Practice exam aesf500 by Adriana Ford-Thompson in Adriana's workspace.
HLMB TEST by Hannah Bentham in Hannah's workspace.
Maths Support: Factorising quadratics by Christian Lawson-Perfect in Maths Support Wiki.
MLC Support: Factorising quadratics by Paraic Treacy in Paraic's workspace.
mathcentre: Diagnostic test - Basic maths by Bill Foster in Bill's workspace.
Bradford: Diagnostic test - Basic maths by Bill Foster in Bill's workspace.
Basic Maths Revision by Gareth Woods in Gareth's workspace.
Solving equations: 1 by Aneela Arshad in Aneela's workspace.
maths skills 1 by cormac murphy in cormac's workspace.

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Bill Foster 55 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

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Solving equations

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Bill Foster 55 years, 2 months ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`