Christopher's copy of Solve a quadratic with the formula

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Chris Graham 6 years, 8 months ago

Created this as a copy of Solve a quadratic with the formula.

Name	Status	Author	Last Modified
Solve a quadratic with the formula	Has some problems	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Solve a quadratic with the a-b-c-formula (the quadratic formula)	draft	Kjell Hansen	10/09/2016 09:57
Solve a quadratic with the a-b-c-formula (the quadratic formula)	draft	Richard Miles	14/09/2016 09:10
Christopher's copy of Solve a quadratic with the formula	draft	Chris Graham	06/08/2018 11:22
Luis's copy of Solve a quadratic with the formula	draft	Luis Hernandez	30/11/2018 01:12
Ugur's copy of Solve a quadratic with the formula	Ready to use	Ugur Efem	01/11/2021 12:31

There are 5 other versions that do you not have access to.

Question statement

Find the roots of the following quadratic equation.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	3
c	integer	1
b	integer	2
d	integer	7
f	integer	6
s3	integer	1
s2	integer	-1
s1	integer	1
n4	number	19
n2	integer	-19
disc	number	361
rdis	string	The roots exist and are distin
n1	integer	23
rep	string
n3	integer	12
c1	integer	1
n5	integer	6
d1	integer	7

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Generated value: `integer`

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Parts

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1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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1. Step Information only

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Solve for $x$ :

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

$x=$ Gap 0. or 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 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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This question is used in the following exam:

Inbbavathie's copy of Practice Questions by Inbbavathie Ravi in Summer Project.

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Christopher's copy of Solve a quadratic with the formula

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Chris Graham 6 years, 8 months ago

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Generated value: `integer`

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Christopher's copy of Solve a quadratic with the formula

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Chris Graham 6 years, 8 months ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`