Simultaneous Equations : Quadratics 4

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Ben McGovern	said	Ready to use	3 years, 8 months ago
Ruth Hand	said	Has some problems	3 years, 9 months ago

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Ben McGovern 2 years, 6 months ago

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Ben McGovern 3 years, 8 months ago

Gave some feedback: Ready to use

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Created this as a copy of Simultaneous Equations : Quadratics 3.

Name	Status	Author	Last Modified
Simultaneous Equations : Quadratics 1	Ready to use	Ben McGovern	06/09/2023 13:28
Simultaneous Equations : Quadratics 2	Ready to use	Ben McGovern	14/09/2022 17:27
Simultaneous Equations : Quadratics 3	Ready to use	Ben McGovern	06/09/2023 13:23
Simultaneous Equations : Quadratics 4	Ready to use	Ben McGovern	14/09/2022 17:28
AC5 Simultaneous Equations (one non-linear)	Ready to use	Andrew Neate	09/04/2024 15:11

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Question statement

The line $\simplify{y={m}x+{c1}}$ touches the curve $\simplify{y={a}x^2+k*x+{c2}}$ . Find the possible values of $k$ .

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			Name	Type	Generated Value

a	integer	1
c1	integer	6
m	integer	-4
c2	integer	7
k1	number	-6
k2	number	-2

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Advice

You are being asked to find the possible values of $k$ such that the line $\simplify{y={m}x+{c1}}$ touches the curve $\simplify{y={a}x^2+k*x+{c2}}$ .

By equating these functions, we can form a quadratic in terms of $x$ :

$\begin{split} \simplify{{a}x^2+k*x+{c2}} &\,= \simplify{{m}x+{c1}} \\ \simplify{{a}x^2+(k-{m})*x+{c2-c1}} &\,=0. \end{split}$

Since we are told that the line touches the curve rather than intercepts it, this is equivalent to this quadratic equation only having one root.

Recall that a quadratic of the form $ax^2+bx+c=0$ only has one root when its discriminant, $b^2-4ac$ , equals $0$ .

So, for

$\simplify{{a}x^2+(k-{m})*x+{c2-c1}} =0,$ the discriminant is

$\simplify{(k-{m})^2 - 4{a}{c2-c1}}.$

By equating this to $0$ , we can form a quadratic in terms of $k$ :

$\begin{split} \simplify{(k-{m})^2 - 4{a}{c2-c1}}&\,=0 \\ \simplify{k^2-{2m}*k+{m^2} - 4{a}{c2-c1}} &\,=0 \\ \end{split}$

By factorising, we can find the possible values of $k$ :

$\begin{split} \simplify{k^2-{2m}*k+{m^2} - 4{a}{c2-c1}} &\,=0 \\ \simplify{(k-{k1})(k-{k2})} &\,= 0 \end{split}$

Therefore,

$k_1 = \var{k1} \quad \text{and} \quad k_2=\var{k2}.$

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Ben McGovern 2 years, 6 months ago

Ben McGovern 3 years, 8 months ago

Ruth Hand 3 years, 9 months ago

Ben McGovern 4 years, 8 months ago

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Simultaneous Equations : Quadratics 4

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Ben McGovern 2 years, 6 months ago

Ben McGovern 3 years, 8 months ago

Ruth Hand 3 years, 9 months ago

Ben McGovern 4 years, 8 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

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Generated value: `integer`