Graphing Linear Equations: Solving x2+4x-7=0 with Hints

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Discussion Overview

The discussion revolves around solving the quadratic equation x² + 4x - 7 = 0 and finding a corresponding linear equation to graph alongside it. Participants seek hints and clarification on how to approach the problem, particularly regarding the nature of the equations involved.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asks for hints on how to start solving the quadratic equation and what linear graph to draw for intersection.
  • Another participant questions what the first graph is, leading to clarification that it is a quadratic graph.
  • There is a suggestion to combine the quadratic equation with a linear equation, but confusion arises about the method of combination.
  • Some participants propose that the equations cannot be solved simultaneously since both are quadratic, while others argue that the second equation is not quadratic.
  • There is a call for participants to simply solve the equations, but skepticism remains about the feasibility of the approach.

Areas of Agreement / Disagreement

Participants express differing views on whether the equations can be solved simultaneously, with some asserting that both are quadratic and thus cannot be combined in that way, while others challenge this assertion. The discussion remains unresolved regarding the correct approach to the problem.

Contextual Notes

There is ambiguity regarding the definitions of the equations and the method of solving them, as well as the assumptions about their forms. The discussion does not clarify the nature of the linear equation needed for the graph.

chikis
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Hello folks, here is a problem. I don't know how and where to start solving it:
A student is told to draw a linear graph on the same axis such that the intersection of the two graph will give the solutions to the equation x2+ 4x -7 =0. What is the equation of the linear graph he needs to draw?
A. x=1 B. x=-1 C. y=1 D. y=-1 E. x+y=1
I don't know where to start from. Can anyone help? Just give me hints that I will follow in proceeding towards the calculation. Please note that I mean x squared when I wrote x2. Thank you.
 
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hello chikis! :smile:
chikis said:
A student is told to draw a linear graph on the same axis such that the intersection of the two graph …

but what's the first graph? :confused:
 
tiny-tim said:
hello chikis! :smile:


but what's the first graph? :confused:

The first graph is a quadratic graph (y=x2+4x-6).
 
chikis said:
The first graph is a quadratic graph (y=x2+4x-6).

ok, so you need to combine y = x2 + 4x - 6 with x2 + 4x - 7 = 0 :wink:
 
tiny-tim said:
ok, so you need to combine y = x2 + 4x - 6 with x2 + 4x - 7 = 0 :wink:

You mean I should combine them by adding it up together as in y= (x2+4x-6) + (x2+4x-7)=0. Is that what you mean?
 
no, i mean combine them as in simultaneous equations :smile:

('cos you want them to be simultaneously true :wink:)
 
tiny-tim said:
no, i mean combine them as in simultaneous equations :smile:

('cos you want them to be simultaneously true :wink:)

You mean I should arrange and solve them like this like if am solving simultanous equation?:
y=x2+4x-6
x2+4x-7=0.
Is that what you mean?
 
yes! :smile:
 
tiny-tim said:
yes! :smile:

That cannot be true because all the two equation resembles quadratic equation. In that case it (the both equation) cannot be solved simultanously. For that to be possible it means that one of the equation has to be linear and the other quadratic.
 
  • #10
chikis said:
That cannot be true because all the two equation resembles quadratic equation. In that case it (the both equation) cannot be solved simultanously.

no, the second equation isn't y = x2 + 4x - 7 :wink:
 
  • #11
tiny-tim said:
no, the second equation isn't y = x2 + 4x - 7 :wink:

does that look like linear equation to you?
 
  • #12
chikis said:
You mean I should arrange and solve them like this like if am solving simultanous equation?:
y=x2+4x-6
x2+4x-7=0.

just solve them!
 
  • #13
tiny-tim said:
just solve them!

That cannot be possible. It can't work that way. Maybe you should try it; let's see what you will get. How about that?
 

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