Solving this quadratic equation on coordinate plane

Phavonic

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Hi,

I am told to solve the quadratic equation given below after plotting the function above it.

1 - x - x2 = y

2x2 - 4x + 1 = 0

But i'm not sure I'm doing it right. I was sure that all I had to do was change the subject of the quadratic formula somewhat, to make the left-hand side of both equations look the same. Thus both will be 1 - x - x2 = the new function.

Either this is a bit difficult to do and I'm not seeing how to do it, or I have got it totally wrong and it's asking me to do something else. Can someone please set me straight. Thanks
 
I am told to solve the quadratic equation given below after plotting the function above it.
1 - x - x2 = y
2x2 - 4x + 1 = 0
But i'm not sure I'm doing it right. I was sure that all I had to do was change the subject of the quadratic formula somewhat, to make the left-hand side of both equations look the same. Thus both will be 1 - x - x2 = the new function. Either this is a bit difficult to do and I'm not seeing how to do it, or I have got it totally wrong and it's asking me to do something else. Can someone please set me straight. Thanks
Is your given task to solve \(\displaystyle 2x^2-4x+1=0~?\). If it is then what does the other function have to do with that task?
It seems that this is a simple quadratic equation to solve. Please reply with explanations.
 
Hi,

I am told to solve the quadratic equation given below after plotting the function above it.

1 - x - x2 = y

2x2 - 4x + 1 = 0

But i'm not sure I'm doing it right. I was sure that all I had to do was change the subject of the quadratic formula somewhat, to make the left-hand side of both equations look the same. Thus both will be 1 - x - x2 = the new function.

Either this is a bit difficult to do and I'm not seeing how to do it, or I have got it totally wrong and it's asking me to do something else. Can someone please set me straight. Thanks

If you could either recite the exact wording of the problem or, better yet, post an image of the problem, it would be easier to understand. I think your interpretation of it has lost some of the meaning. That is to say, by stating what you think it is asking, you deprive others of seeing exactly how it reads.

Is it asking you to solve for where the two curves intersect?
 
If you could either recite the exact wording of the problem or, better yet, post an image of the problem, it would be easier to understand. I think your interpretation of it has lost some of the meaning. That is to say, by stating what you think it is asking, you deprive others of seeing exactly how it reads.

Is it asking you to solve for where the two curves intersect?

Yes, it's asking me to solve x in the quadratic equation where both lines intersect (that is, with the line of the equation that's produced from the two given). I assumed this was the only meaning of the question.

Put exactly it says....

Question 10: Plot the following function f(x)---> 1 - x - x2

Question 14: Using your graph for question 10, plot a suitable straight line which enables you to solve the equation 2x2 - 4x +1 = 0

Previous questions have for example asked me to plot a line for

x2 + 2x = y

And then with this to solve the equation

x2 + 3x = 0

So I change the formula here to make the left-side the same as the first equation

x2 + 2x = -x

So if x2 + 2x = y and x2 + 2x = -x, then y = -x. I am drawing this second line y = -x and where it crosses the first line (y = x2 + 2x) will be the values for x in x2 +3x= 0 .

I'm applying this method to questions 10 and 14 given above, but it doesn't seem to work because of the negative or positive values of the x's. So I wasn't sure if I am getting something wrong in changing the formula, or is the question asking me something different; but nothing in my text book hints at the latter. Any ideas?
 
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Hi,

I am told to solve the quadratic equation given below after plotting the function above it.

1 - x - x2 = y

2x2 - 4x + 1 = 0

But i'm not sure I'm doing it right. I was sure that all I had to do was change the subject of the quadratic formula somewhat, to make the left-hand side of both equations look the same. Thus both will be 1 - x - x2 = the new function.

Either this is a bit difficult to do and I'm not seeing how to do it, or I have got it totally wrong and it's asking me to do something else. Can someone please set me straight. Thanks
2x2 - 4x + 1 = 0 ==>x2 = 2x - 0.5. Make this substitution in the 1st equation and you'll get a linear equation. What exactly are you confused about?
 
2x2 - 4x + 1 = 0 ==>x2 = 2x - 0.5. Make this substitution in the 1st equation and you'll get a linear equation. What exactly are you confused about?

Well, if you read my follow up you can find out for yourself just what I'm "confused about", that's your job and the point of this forum, not to be smarmy. I explained the method I was using and asked if there is another method. You're assuming that people who need help with maths are supposed to know all the answers already!
 
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Well, if you read my follow up you can find out for yourself just what I'm "confused about", that's your job and the point of this forum, not to be smarmy. I explained the method I was using and asked if there is another method.

It's very helpful that we now have the actual wording of the problem, and an example of what is expected.

We could help you much more easily if you showed your actual work for this problem. Can you do so? Then we can see if we interpret the problem differently than you, or if you just made a simple mistake.

I think Jomo either suggested a different method (not what you were told to do), or is interpreting the problem differently than I do. Again, seeing your work will help.
 
2x2 - 4x + 1 = 0 ==>x2 = 2x - 0.5. Make this substitution in the 1st equation and you'll get a linear equation. What exactly are you confused about?
Well, if you read my follow up you can find out for yourself just what I'm "confused about", that's your job and the point of this forum, not to be smarmy. I explained the method I was using and asked if there is another method. You're assuming that people who need help with maths are supposed to know all the answers already!
@Phavonic, I think we all here understand your frustration at this question. Frankly I am totally confused by it and taught mathematics for fifty years. It seems that whoever wrote this questing has a very specific objective in mind. We like you do not have access to that mind. So we are doing our job here thank you very much.
 
Well, if you read my follow up you can find out for yourself just what I'm "confused about", that's your job and the point of this forum, not to be smarmy. I explained the method I was using and asked if there is another method. You're assuming that people who need help with maths are supposed to know all the answers already!
Perhaps if you had followed the forum's guidelines,

https://www.freemathhelp.com/forum/threads/112086-Guidelines-Summary?p=433156&viewfull=1#post433156

you would have got prompter and better help. Did you see where it says to show your work even if you think it is wrong? Did you do that? NO! Did you see where it says to tell us where the problem came from so we have some clue as to its context? Did you do that? NO! Or perhaps you failed to read the guidelines at all.

In any case, you are asking for "free" help. The volunteers here do not work at a "job." No one pays us, certainly not you. So don't give yourself the air of the lord of the manner looking down his nose at a parlor maid whom he can sack for not showing proper deference to her betters.
 
Perhaps if you had followed the forum's guidelines,

https://www.freemathhelp.com/forum/threads/112086-Guidelines-Summary?p=433156&viewfull=1#post433156

you would have got prompter and better help. Did you see where it says to show your work even if you think it is wrong? Did you do that? NO! Did you see where it says to tell us where the problem came from so we have some clue as to its context? Did you do that? NO! Or perhaps you failed to read the guidelines at all.

In any case, you are asking for "free" help. The volunteers here do not work at a "job." No one pays us, certainly not you. So don't give yourself the air of the lord of the manner looking down his nose at a parlor maid whom he can sack for not showing proper deference to her betters.


I've read the rules and guidelines, but I like I said, I assumed that you as mathematicians had understood the question, if you don't then I can't know how it is that you're not understanding it. Like pka intimated, I don't have access to your minds, and you don't have access to my mind,—the problem of this world! with much hubris and impatience besides. I am just a beginner who is working from crumbs, so I don't know what your overview is in this complex world of mappings and functions; all I have is one chapter on this subject from a beginners' A-level textbook.


You asked me if I had shown my work,even if I think it is wrong (and angrily exclaimed NO!). But I disagree and think I did this as best I could by showing a "previous question" that I'd gotten correct (where it gave me x2 + 2x = y and x2 + 3x = 0), and where I said I was trying to apply this same method in changing the formula of 2x2 - 4x + 1 = 0 so that the left-hand side looks the same as the first equation, as per the sum I got correct, but I couldn't do this because of the positive and negative values of x don't hold out. Am I so wrong in assuming that you would intuitively know what this means without actually writing it down?


But as messy as this will look, here it is...


I want to change 2x2 - 4x + 1 = 0 to 1 - x - x2 = ??? by moving as much as I can to the right-hand like of the = sign.


2x – 4x + 1 = 0
2x – x +1 = 3x
x – x + 1 = x/2


And here I stop because of the positive and negative values of x can't get me to 1 - x - x2 on the left-hand side. Since I also assumed this method is wrong, I thought it wasn't worth writing it out in full.


You also asked me to state where the problem came from so you can see its context, and again got angry.But likewise I assumed you would know intuitively its context since it is so basic (and where it comes from seems erroneous).


But as a further a clearing up, my textbook gives only two examples before going into the questions (this is after the basics of mapping and functions).


It asks to solve x2 - 1/x = x3


And asks to plot two functions on a graph


y = x2 – 1/x
y = x3


And says the points of intersection of the two lines will solve x.


The second example says solve x3 + 2x -3 = 0 by plotting a line for x3 + 2x = y, and then change the left-hand side of the quadratic formula to look like the left-hand side of this formula which makes the line. So....


x3 + 2x - 3 = 0
x3 + 2x = 3


Since x3 + 2x = 3, and x3 + 2x = y, then y = 3, and then plotting of this line will run dissect through the line of x3 + 2x = y and show the two points which solve the quadratic. After this the questions are quite easy, but then it gets to the question that I had quoted verbatim. And so, you see, all I was trying to do was change the subject of the formula for the left-hand side of the quadratic...which you see I couldn't do, and so brought me to you kind people.


When I ask this question I get a lot of confusion back, I am even more confused and have no idea how it cannot be clear. So, as ever I'm just asking what method I am supposed to use to solve 2x2 - 4x + 1 = 0 after plotting the line 1 - x - x2 = y and how I am to change the subject of the formula accordingly.


I used the word “job” rhetorically because I was asked what I am “confused about” rhetorically; and you have to admit that you are not volunteers in the broad sense of that word, you are also people who love maths and puzzles, no?


You know I can never ask questions on this forum again if you gang up to me and make me out to be the wrong one. But it's just a problem of communicating and patience.
 
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I've read the rules and guidelines, but I like I said, I assumed that you as mathematicians had understood the question, if you don't then I can't know how it is that you're not understanding it. Like pka intimated, I don't have access to your minds, and you don't have access to my mind,—the problem of this world! with much hubris and impatience besides. I am just a beginner who is working from crumbs, so I don't know what your overview is in this complex world of mappings and functions; all I have is one chapter on this subject from a beginners' A-level textbook.


You asked me if I had shown my work,even if I think it is wrong (and angrily exclaimed NO!). But I disagree and think I did this as best I could by showing a "previous question" that I'd gotten correct (where it gave me x2 + 2x = y and x2 + 3x = 0), and where I said I was trying to apply this same method in changing the formula of 2x2 - 4x + 1 = 0 so that the left-hand side looks the same as the first equation, as per the sum I got correct, but I couldn't do this because of the positive and negative values of x don't hold out. Am I so wrong in assuming that you would intuitively know what this means without actually writing it down?


But as messy as this will look, here it is...


I want to change 2x2 - 4x + 1 = 0 to 1 - x - x2 = ??? by moving as much as I can to the right-hand like of the = sign.


2x – 4x + 1 = 0
2x – x +1 = 3x
x – x + 1 = x/2


And here I stop because of the positive and negative values of x can't get me to 1 - x - x2 on the left-hand side. Since I also assumed this method is wrong, I thought it wasn't worth writing it out in full.


You also asked me to state where the problem came from so you can see its context, and again got angry.But likewise I assumed you would know intuitively its context since it is so basic (and where it comes from seems erroneous).


But as a further a clearing up, my textbook gives only two examples before going into the questions (this is after the basics of mapping and functions).


It asks to solve x2 - 1/x = x3


And asks to plot two functions on a graph


y = x2 – 1/x
y = x3


And says the points of intersection of the two lines will solve x.


The second example says solve x3 + 2x -3 = 0 by plotting a line for x3 + 2x = y, and then change the left-hand side of the quadratic formula to look like the left-hand side of this formula which makes the line. So....


x3 + 2x - 3 = 0
x3 + 2x = 3


Since x3 + 2x = 3, and x3 + 2x = y, then y = 3, and then plotting of this line will run dissect through the line of x3 + 2x = y and show the two points which solve the quadratic. After this the questions are quite easy, but then it gets to the question that I had quoted verbatim. And so, you see, all I was trying to do was change the subject of the formula for the left-hand side of the quadratic...which you see I couldn't do, and so brought me to you kind people.


When I ask this question I get a lot of confusion back, I am even more confused and have no idea how it cannot be clear. So, as ever I'm just asking what method I am supposed to use to solve 2x2 - 4x + 1 = 0 after plotting the line 1 - x - x2 = y and how I am to change the subject of the formula accordingly.


I used the word “job” rhetorically because I was asked what I am “confused about” rhetorically ; and you have to admit that you are not volunteers in the broad sense of that word, you are also people who love maths and puzzles, no?


You know I can never ask questions on this forum again if you gang up to me and make me out to be the wrong one. But it's just a problem of communicating and patience.
I want to change 2x2 - 4x + 1 = 0 to 1 - x - x2 = ???

Since 2x2 - 4x + 1 = 0 we have 2x2 = 4x -1 and then dividing by 2 we get x2 = (4x -1)/2 = 2x - 0.5 --note that I solved for x2 because x2 is what the other equation has.

Now y =
1 - x - x2 = 1 -x -(2x - 0.5) = -3x + 1.5.

So
y = 1 - x - x2 = -3x + 1.5

If I understand your question, and I am not sure that I do, the answer is above.

EDIT: I fixed my arithmetic above.
 
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this complex world of mappings and functions; all I have is one chapter on this subject from a beginners' A-level textbook. (and angrily exclaimed NO!). But I disagree and think I did this as best I could by showing a "previous question" that I'd gotten correct (where it gave me x2 + 2x = y and x2 + 3x = 0)
I want to change 2x2 - 4x + 1 = 0 to 1 - x - x2 = ??? by moving as much as I can to the right-hand like of the = sign.
You know I can never ask questions on this forum again if you gang up to me and make me out to be the wrong one. But it's just a problem of communicating and patience.
You know don't you that no one here angrily exclaimed anything to you. You seem to be projecting an attitude that is simply not there.
So read this carefully.
At least two PhD mathematicians tried to say to you that the question as posted seems to have grown out of a series of other graphing questions designed to answer a given problem. As I posted: I can see no relation that the graph of \(\displaystyle 1-x-x^2\) can have in the way of solving \(\displaystyle 2x^2-4x+1=0\). That said, someone asked you if the task were to solve \(\displaystyle 2x^2-4x+1=1-x-x^2\) or where the two intersect? Did you answer and/or try to clarify/simplify our understanding?
I think not. You are welcome here only if you can be civil and post clearly defined problems.
 
I want to change 2x2 - 4x + 1 = 0 to 1 - x - x2 = ???

Since 2x2 - 4x + 1 = 0 we have 2x2 = 4x -1 and then dividing by 2 we get x2 = (4x -1)/2 = 2x - 0.5 --note that I solved for x2 because x2 is what the other equation has.

Now y =
1 - x - x2 = 1 -x -(2x - 0.5) = -x + 1.5.

So
y = 1 - x - x2 = -x + 1.5

If I understand your question, and I am not sure that I do, the answer is above.
Here's my take on the problem.

\(\displaystyle 1-x-x^2 = y\)
\(\displaystyle -x^2-x+1=y\) …. (*)

\(\displaystyle 2x^2-4x+1=0\)
\(\displaystyle x^2 - 2x +0.5=0\) ...dividing both sides by 2
\(\displaystyle -x^2+2x-0.5 =0\) … multiplying both sides by -1
\(\displaystyle -x^2 -x +1 = -3x + 1.5 \) .... adding -3x+1.5 to each side to achieve the same LHS as at (*)

Now equating RHSs,
y = -3x + 1.5

So you can find the solution to the quadratic equation \(\displaystyle 2x^2-4x+1=0\) by finding where this line interests the graph of the parabola you have drawn.

(BTW, I don't know why someone would ask you to do this - never before seen a question like it. There are much more straight-forward ways of solving a quadratic equation!)
 
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I want to change 2x2 - 4x + 1 = 0 to 1 - x - x2 = ???

Since 2x2 - 4x + 1 = 0 we have 2x2 = 4x -1 and then dividing by 2 we get x2 = (4x -1)/2 = 2x - 0.5 --note that I solved for x2 because x2 is what the other equation has.

Now y =
1 - x - x2 = 1 -x -(2x - 0.5) = -x + 1.5. I think this should be -3x+15

So
y = 1 - x - x2 = -x + 1.5

If I understand your question, and I am not sure that I do, the answer is above.
see comment
 
I've read the rules and guidelines, but I like I said, I assumed that you as mathematicians had understood the question, if you don't then I can't know how it is that you're not understanding it. Like pka intimated, I don't have access to your minds, and you don't have access to my mind,—the problem of this world! with much hubris and impatience besides. I am just a beginner who is working from crumbs, so I don't know what your overview is in this complex world of mappings and functions; all I have is one chapter on this subject from a beginners' A-level textbook.

You asked me if I had shown my work,even if I think it is wrong (and angrily exclaimed NO!). But I disagree and think I did this as best I could by showing a "previous question" that I'd gotten correct (where it gave me x2 + 2x = y and x2 + 3x = 0), and where I said I was trying to apply this same method in changing the formula of 2x2 - 4x + 1 = 0 so that the left-hand side looks the same as the first equation, as per the sum I got correct, but I couldn't do this because of the positive and negative values of x don't hold out. Am I so wrong in assuming that you would intuitively know what this means without actually writing it down?

It's unfortunate that on this sort of site, you can get people piling up on you saying the same thing in different ways, so that even if they are not being rude, it can feel that way.

But perhaps it is becoming clear to you that we are all confused, because this is not a standard problem that we are all familiar with. We really need the information we have asked for, in order to understand what in the world you are being asked to do. And we really don't know what you are thinking without being told, because we aren't coming at this with your particular background (nor do we have a clear idea what your background is). You have gradually given us the information we need for both. Thank you. Now let's all try to calm down.

But as messy as this will look, here it is...

I want to change 2x2 - 4x + 1 = 0 to 1 - x - x2 = ??? by moving as much as I can to the right-hand like of the = sign.

2x2 – 4x + 1 = 0
2x2 – x +1 = 3x
x2 – x + 1 = x/2

You have made a fundamental mistake here, though I'm not quite sure what you were thinking. You have divided the first term on the left by 2, but somehow divided the whole right side by 6. Or maybe I'm wrong in the way I put the exponents where I think you intended them.

So you may be trying to do the right thing, or you might not. And I'm not entirely sure what the right thing is in the first place.

And here I stop because of the positive and negative values of x can't get me to 1 - x - x2 on the left-hand side. Since I also assumed this method is wrong, I thought it wasn't worth writing it out in full.

You also asked me to state where the problem came from so you can see its context, and again got angry.But likewise I assumed you would know intuitively its context since it is so basic (and where it comes from seems erroneous).

But as a further a clearing up, my textbook gives only two examples before going into the questions (this is after the basics of mapping and functions).

It asks to solve x2 - 1/x = x3

And asks to plot two functions on a graph

y = x2 – 1/x
y = x3

And says the points of intersection of the two lines will solve x.

This example is a standard concept, and is what I would have thought your problem would expect -- except that it clearly doesn't. There is no way I would ever try to solve 2x2 - 4x + 1 = 0 graphically by graphing y = 1 - x - x2 and some line. I'd graph y = 2x2 - 4x + 1, or y = 2x2 - 4x.

The second example says solve x3 + 2x -3 = 0 by plotting a line for x3 + 2x = y, and then change the left-hand side of the quadratic formula to look like the left-hand side of this formula which makes the line. So....

x3 + 2x - 3 = 0
x3 + 2x = 3

Since x3 + 2x = 3, and x3 + 2x = y, then y = 3, and then plotting of this line will run dissect through the line of x3 + 2x = y and show the two points which solve the quadratic. After this the questions are quite easy, but then it gets to the question that I had quoted verbatim. And so, you see, all I was trying to do was change the subject of the formula for the left-hand side of the quadratic...which you see I couldn't do, and so brought me to you kind people.

This, too, is something that looks entirely familiar. The problem you are asking about seems too different from these to be sure what they want you to do. Possibly, somewhere in the "quite easy" part you have omitted, there is a hidden clue ...

When I ask this question I get a lot of confusion back, I am even more confused and have no idea how it cannot be clear. So, as ever I'm just asking what method I am supposed to use to solve 2x2 - 4x + 1 = 0 after plotting the line 1 - x - x2 = y and how I am to change the subject of the formula accordingly.

Let me show you my own idea of what you are supposed to do, which may or may not be "correct".

As you did, I would be trying to somehow rewrite 2x2 - 4x + 1 = 0 to use 1 - x - x2, that is, -x2 - x + 1. So I'd start by multiplying everything by -1/2, in order to get the correct coefficient of the squared term:

2x2 - 4x + 1 = 0 ---> -x2 + 2x - 1/2 = 0

Note that I have to multiply every term, not just one term.

Now I want to change the middle term to -x; to do that I have to subtract 3x from both sides:

-x2 - x - 1/2 = -3x

Now I want to change the -1/2 to +1; so I add 3/2 to both sides:

-x2 - x + 1 = -3x + 3/2

So I can solve the equation by intersecting y = -x2 - x + 1 with y =
-3x + 3/2.

Again, this is far from an efficient way to solve it, so I'm hoping that if I saw the rest of the material between the examples and the exercises, I might see some different trick that they are somehow hinting at.
 
And what will that do?
Will give us an indication of "where this guy is at"!
If he can't solve a quadratic, then we can simply wave him bye-bye...
In a polite fashion, of course :rolleyes:
 
At least two PhD mathematicians tried to say to you that the question as posted seems to have grown out of a series of other graphing questions designed to answer ...
Then stating this is not part of a valid argument. If anything, I would not
trust their points of view as much, because they would tend to be more
out-of-touch and arrogant with lower level students. The helpers to trust more are
the tutors who regularly teach and tutor students at the intermediate
mathematics level with quadratic equations.
 
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