How to proceed with this?

[allegansveritatem]

Problem:
I have done a lot of fooling with this but I will only append one example:

This is a hard one, a kind of Gordian knot of a problem.Or so it seems right now.

 

[Dr.Peterson]

What is your goal in step 4? It doesn't lead anywhere.

Look at step 3. You have a quadratic equation, [MATH](2)x^2 + (2b)x + (b^2 - 4) = 0[/MATH]. How do you determine how many solutions a quadratic equation has? (Hint: the discriminant is Alexander's sword.)

 

[pka]

Problem:View attachment 21133

The first equation \(x^2+y^2=4\) is a circle centred at the origin with radius two.
The slope one. SEE I also SEEII

 

[HallsofIvy]

Yes, the graph of \(\displaystyle x^2+ y^2= 4\) is a circle with center at the origin and radius 2. And y= x+ b is a line with slope 1 and y-intercept b. A line
1) crosses through the circle and crosses it twice
2) is tangent to the circle and touches it once.
3) completely misses the circle so never touches or crosses it.

At any (x, y) point where the line touches or crosses the circle, y is the same in both equation so we can write the first as \(\displaystyle x^2+ y^2= x^2+ (x+ b)^2= x^2+ x^2+ 2bx+ b^2= 2x^2+ 2bx+ b^2= 4\).
\(\displaystyle 2x^2+ 2bx+ (b^2- 4)= 0\).

That is a quadratic equation and roots of the quadratic equation, \(\displaystyle ax^2+ bx+ c= 0\) (not the same "b" as above, of course) are are given by \(\displaystyle x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\). There will be two real roots if the "discriminant", \(\displaystyle b^2- 4ac\), is positive, one root if it is 0, and no real root if it is negative.

Here a= 2, b= 2b, and \(\displaystyle c= b^2- 4\) so the discriminant is \(\displaystyle 4b^2- 8b^2+ 32= -4b^2- 32\). Now, for what values of b is that positive, negative, or 0?

 

[Steven G]

Of course you got lost after you lost an equal sign in (4) and (5).

 

[allegansveritatem]

What is your goal in step 4? It doesn't lead anywhere.

Look at step 3. You have a quadratic equation, [MATH](2)x^2 + (2b)x + (b^2 - 4) = 0[/MATH]. How do you determine how many solutions a quadratic equation has? (Hint: the discriminant is Alexander's sword.)

As I look at it in the cold light of morning I can only say that with step four I was trying to factor step three and thus zero out a couple of x's. I will take your hint and check out the discriminant--which concept I will have to review. As I recall it was some element of the quadratic equation, the part sheltered under the radical. I will see what I can see and get back.

 

[allegansveritatem]

The first equation \(x^2+y^2=4\) is a circle centred at the origin with radius two.
The slope one. SEE I also SEEII

Thanks for links. I wouldn't have thought of that. I should have recognized that equation as a unit circle.

 

[allegansveritatem]

Yes, the graph of \(\displaystyle x^2+ y^2= 4\) is a circle with center at the origin and radius 2. And y= x+ b is a line with slope 1 and y-intercept b. A line
1) crosses through the circle and crosses it twice
2) is tangent to the circle and touches it once.
3) completely misses the circle so never touches or crosses it.

At any (x, y) point where the line touches or crosses the circle, y is the same in both equation so we can write the first as \(\displaystyle x^2+ y^2= x^2+ (x+ b)^2= x^2+ x^2+ 2bx+ b^2= 2x^2+ 2bx+ b^2= 4\).
\(\displaystyle 2x^2+ 2bx+ (b^2- 4)= 0\).

That is a quadratic equation and roots of the quadratic equation, \(\displaystyle ax^2+ bx+ c= 0\) (not the same "b" as above, of course) are are given by \(\displaystyle x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\). There will be two real roots if the "discriminant", \(\displaystyle b^2- 4ac\), is positive, one root if it is 0, and no real root if it is negative.

Here a= 2, b= 2b, and \(\displaystyle c= b^2- 4\) so the discriminant is \(\displaystyle 4b^2- 8b^2+ 32= -4b^2- 32\). Now, for what values of b is that positive, negative, or 0?

Thanks for the reply. I will check it out later when I get into math mind and see what I can do with it.

 

[allegansveritatem]

Of course you got lost after you lost an equal sign in (4) and (5).

I was locked onto the pursuit of factors and lost sight of why I was seeking them--tunnel vision math.

 

[allegansveritatem]

Yes, the graph of \(\displaystyle x^2+ y^2= 4\) is a circle with center at the origin and radius 2. And y= x+ b is a line with slope 1 and y-intercept b. A line
1) crosses through the circle and crosses it twice
2) is tangent to the circle and touches it once.
3) completely misses the circle so never touches or crosses it.

At any (x, y) point where the line touches or crosses the circle, y is the same in both equation so we can write the first as \(\displaystyle x^2+ y^2= x^2+ (x+ b)^2= x^2+ x^2+ 2bx+ b^2= 2x^2+ 2bx+ b^2= 4\).
\(\displaystyle 2x^2+ 2bx+ (b^2- 4)= 0\).

That is a quadratic equation and roots of the quadratic equation, \(\displaystyle ax^2+ bx+ c= 0\) (not the same "b" as above, of course) are are given by \(\displaystyle x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\). There will be two real roots if the "discriminant", \(\displaystyle b^2- 4ac\), is positive, one root if it is 0, and no real root if it is negative.

Here a= 2, b= 2b, and \(\displaystyle c= b^2- 4\) so the discriminant is \(\displaystyle 4b^2- 8b^2+ 32= -4b^2- 32\). Now, for what values of b is that positive, negative, or 0?

seems to me that -4b^2 -32 can't yield any real solutions.

 

[allegansveritatem]

I worked on this for over an hour today but have no images to present ...my camera seems to have gotten itself misplaced. I can summarize my results thus: I transformed the first equation into y= the sqrt of 4-x^2, graphed that and discovered that x and y both 2 and that would mean that the discriminant should be such as to indicate that there are two real solutions that are the same. But I didn't come up with any such discriminant. Mine is/was -4b^2-32. How do you get a zero out of that?

 

[JeffM]

I think you are missing the point of the exercise..

[MATH]x^2 + y^2 = 4 \text { and } y = x + b \implies \\ x^2 + (x + b)^2 = 4 \implies 2x^2 + 2bx + b^2 - 4 = 0 \implies \\ x = \dfrac{-2b \pm \sqrt{4b^2 - 4(2)(b^2 - 4)}}{2 * 2} = \dfrac{-2b \pm 2\sqrt{b^2 - 2b^2 + 8}}{2 * 2} = \dfrac{-b \pm\sqrt{8 - b^2}}{2}.[/MATH]Why?

So under what circumstance will x have just one real value? And two?

 

[Steven G]

Here is how I would do this problem.
The circle is centered at the origin. The line y = x + 0 will cross the circle twice.
If we let b differ from 0 by just a little, then the line will cross the circle twice.
If we let b differ from 0 by a bit more then the line will cross the circle twice.
At some point if b differs from 0 enough then the line will cross the circle in exactly one point.

Now we need to remove words like enough, a little and a bit more!

The slope of all lines of the form y=x+b is 1.

All we need to do is find out the two points where the slope of the tangent line to the circle is 1. Where does these two tangent lines cross the y-axis. Those are your two b* values. If b (in y=x+b) is in-between the two b* values you just found the line will cross the circle twice. If b equals one of the two b*values you just found then the line will cross the circle in one place.

 

[allegansveritatem]

well, I went back at this today and keeping in mind the posts in this thread came up with this:

they ask for a)one solution b )two solutions and c) no solutions. I did not deal with c but I would just enter b= more than 8

 

[Steven G]

It does not follow that if y^2=4-x^2 then y = sqrt(4-x^2). Then you replaced y with x-b? Why?

Please try again.

 

[allegansveritatem]

It does not follow that if y^2=4-x^2 then y = sqrt(4-x^2). Then you replaced y with x-b? Why?

Please try again.

no, it does not...but I was looking at the discriminant and the question was what were the values of b that would allow for one solution, for two and for none. The values I gave would, it seemed at the time at least, satisfy those conditions. Yes, I will try again and get back.

 

[JeffM]

On your next to last line, you correctly have this root as being relevant.

[MATH]\sqrt{-4b^2 + 32}.[/MATH]
Because I am lazy, I'd simplify it to

[MATH]\sqrt{-4b^2 + 32} = \sqrt{4(8 - b^2)} = 2\sqrt{8 - b^2}.[/MATH]
It is not necessary to simplify, but it does make it easier to think.

Now is it really true that [MATH]8 - b^2 \ge 0 \implies 8 \ge b?[/MATH]
So you got almost to the end before making a careless error and a more subtle one.

I am really saying much the same thing as jomo.

 

[Steven G]

Please read post #13 and see if it makes any sense to you. If it does then give it a try, maybe you'll like it!

 

[allegansveritatem]

Here is how I would do this problem.
The circle is centered at the origin. The line y = x + 0 will cross the circle twice.
If we let b differ from 0 by just a little, then the line will cross the circle twice.
If we let b differ from 0 by a bit more then the line will cross the circle twice.
At some point if b differs from 0 enough then the line will cross the circle in exactly one point.

Now we need to remove words like enough, a little and a bit more!

The slope of all lines of the form y=x+b is 1.

All we need to do is find out the two points where the slope of the tangent line to the circle is 1. Where does these two tangent lines cross the y-axis. Those are your two b* values. If b (in y=x+b) is in-between the two b* values you just found the line will cross the circle twice. If b equals one of the two b*values you just found then the line will cross the circle in one place.

OK. I will give this a go tomorrow. I did some more work on this today and realized that I have a hazy understanding of the concepts and the procedures involved in this matter of systems of equations. So I think I will go over the chapter again too. I have used systems of equations while studying algebra but those systems were simpler and did not involve more than one solution for each variable.

 

[allegansveritatem]

On your next to last line, you correctly have this root as being relevant.

[MATH]\sqrt{-4b^2 + 32}.[/MATH]
Because I am lazy, I'd simplify it to

[MATH]\sqrt{-4b^2 + 32} = \sqrt{4(8 - b^2)} = 2\sqrt{8 - b^2}.[/MATH]
It is not necessary to simplify, but it does make it easier to think.

Now is it really true that [MATH]8 - b^2 \ge 0 \implies 8 \ge b?[/MATH]
So you got almost to the end before making a careless error and a more subtle one.

I am really saying much the same thing as jomo.

right, simplified is beautiful, to paraphrase somebody. I will have to look at it again. I worked it out again today and got strung out on something to do with the variable (b) and had a photo I was going to upload and a question but then I realized my mistake and so nothing needs be said on the matter. I still haven't clearly understood on a procedural level how to do these systems problems where there is more than one solution per variable. I need to go back and study the chapter again.