why are we interested in only those values of x for which y is zero?

[PG1995]

Hi

I understand that when roots a quadratic equation is found, we are essentially finding values of x for which y or f(x) is zero. But what how does it help us? I mean why are we only interested in values of x which makes y zero. e.g. quadratic formula for finding roots could only be used for finding values of x for which y is zero but it won't help us if we want to know values of x for which y is 2. Will it? Likewise, while solving differential equations we are only interested in those values which make the differential function equal to zero.

**broken link removed**

Regards
PG

 

[MrAl]

Hello again PG,

Well you assumed that we can only solve for y=0 but actually this kind of method is much more general method than that. You only need to see a few more examples to see how versatile this really is. Of course i have an example for you

We start with the quadratic:
y=ax^2+bx+c

where a, b, and c are constants and i did not show some multiplication signs for simplicity in appearance.

To find out the value of x when y is zero we substitute y=0 and we get:
0=ax^2+bx+c

and when we solve for x we get two values for x where y is zero.

But what is to stop us from using any other value of y? For your second thought we might make y=2 as you said, then we would have:
2=ax^2+bx+c

And now we have four constants, but we can reduce this again to three by allowing 'c' to absorb the '2' as simply as subtracting 2 from both sides:
0=ax^2+bx+(c-2)

And now we have again a quadratic, except now when we solve for x we get the values that cause y to be 2.

But the versatility does not stop there. Suppose we dont actually have a constant value of y, but instead we have a simple linear straight line that angles from the horizontal:
y=D*x

That's a sloping line. Now say we want the solution to x when y meets that line. We start again wth:
y=ax^2+bx+c

but now we also have:
y=Dx

so we can make the subts again into the starting quadratic:
Dx=ax^2+bx+c

and now we allow the bx to absorb the Dx (we just subtract Dx from both sides):
0=ax^2+(b-D)x+c

Now this is just a quadratic again, and when we solve it this time we end up with the solutions for the intersection of the parabola and the straight sloping line.

There are a few catches, for example if we try to solve for y=-2 instead of +2 for a parabola that never goes below zero, we will end up with no real solution.

Of course we can also use two quadratics and ask the question if they intersect:
y=ax^2+bx+c
y=Ax^2+Bx+C

and after equating the two we get:
0=(a-A)x^2+(b-B)x+(c-C)

and the solutions (if any) will be the intersections of the two parabolas.

The quadratic is limited only in the order of the equation, that it has to be 2. Thus if we have the two:
y=ax^2+bx+c
y=Ax^3+Bx^2+Cx+D

this no longer admits a quadratic solution, because of the third degree power, so this becomes a third degree problem where we cant use the quadratic formula (directly) anymore. There are ways to solve the third degree too though using methods for third degree equations.

As to the practical applications for this kind of thing, they are endless in number.


From all your questions you've asked here on ET it appears that you could benefit greatly from a short class (or self study) in precalculus before or during a study in calculus, and a little analytic geometry wouldnt hurt either. You seem to be doing quite well with your calculus though (and really great with your English BTW) but you might find it easier to get through with these few prerequisites.

 

[Sceadwian]

Perspective.... You always have to have a perspective.

 

[PG1995]

Thanks a lot, MrAl, for your detailed explanation. It was very useful. When I made that post I was also working on my own to teach myself what happens behind the scenes what equations are solved simultaneously.

By the way, I think the embedded graph in my first post is incorrect. The roots for y=x^2 - 3x - 10^2 are: x = 11.6 or x = -8.6. The roots shown in the graph are: 5 and -2.

Best wishes
PG

 

[MrAl]

Hi,

Yes the roots are 11.612 and -8.612 approximately for the equation x^2-3*x-100, but for the blue curve drawn in your picture the roots shown are correct.

 

[PG1995]

Hi,

Yes the roots are 11.612 and -8.612 approximately for the equation x^2-3*x-100, but for the blue curve drawn in your picture the roots shown are correct.

The equation, say Eq.1, x^2-3*x-100 has roots 11.612 and -8.612. The curve in my first post above is supposedly for equation, Eq.2, x^2-3*x-10^2. The only difference between Eq.1 and Eq.2, in my opinion, is that Eq.2 has written "100" in form "10^2". The roots shown for the curve in my first post, x=5, or x=-2, are for the equation x^2-3*x-10, not for x^2-3*x-100 or x^2-3*x-10^2. Do I make any sense? Please let me know.

Best regards
PG

 

[MrAl]

Hi,

Yes, and i assumed that the curve drawn was a different curve, other than the equation you were actually working with. No big deal really.

 

[PG1995]

Hi,

Yes, and i assumed that the curve drawn was a different curve, other than the equation you were actually working with. No big deal really.

Thanks, MrAl. It would be a big deal from my point of view if I were wrong because it would mean there is gap in my understanding. Now you have confirmed it, now it's no big deal!

Best wishes
PG