Solve a quadratic with the a-b-c-formula (the quadratic formula)

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Richard Miles

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Richard Miles 8 years, 6 months ago

Created this as a copy of Solve a quadratic with the a-b-c-formula (the quadratic 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

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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	3
c	integer	5
b	integer	4
d	integer	1
f	integer	12
s3	integer	1
s2	integer	-1
s1	integer	-1
n4	number	17
n2	integer	17
disc	number	289
rdis	string	The roots exist and are distin
n1	integer	23
rep	string
n3	integer	24
c1	integer	5
n5	integer	12
d1	integer	1

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Generated value: `integer`

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d
disc
f
n1
n2
n3
n5

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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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Solve a quadratic with the a-b-c-formula (the quadratic formula)

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Richard Miles 8 years, 6 months ago

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Solve a quadratic with the a-b-c-formula (the quadratic formula)

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Richard Miles 8 years, 6 months ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`