The roots of a quadratic equation

New member

I cannot figure out how to work out the last part … I figured out that one real solution means real and equal roots (discriminant = 0) but I cannot understand how it is possible to prove that c^2 is between 0 and 1 inclusive.

Dr.Peterson

Elite Member
View attachment 17365

I cannot figure out how to work out the last part … I figured out that one real solution means real and equal roots (discriminant = 0) but I cannot understand how it is possible to prove that c^2 is between 0 and 1 inclusive.
If $$\displaystyle c^2 = 1 - \left(\frac{a+b}{a-b}\right)^2$$, can $$\displaystyle c^2$$ ever be greater than 1? That part should be easy. (By the way, they told you that $$\displaystyle c \ne 1$$, so the "inclusive" part was not really necessary.)

And it should be even easier to see that $$\displaystyle c^2 > 0$$.

If you don't see this, please show your work in attempting it, so we can see if you are going in the wrong direction.

New member
If $$\displaystyle c^2 = 1 - \left(\frac{a+b}{a-b}\right)^2$$, can $$\displaystyle c^2$$ ever be greater than 1? That part should be easy. (By the way, they told you that $$\displaystyle c \ne 1$$, so the "inclusive" part was not really necessary.)

And it should be even easier to see that $$\displaystyle c^2 > 0$$.

If you don't see this, please show your work in attempting it, so we can see if you are going in the wrong direction.

If you are subtracting from 1, the value should be less than 1. Also, yes anything square is positive, so the answer can never be less than 0. But does this involve any working? How can I represent my thoughts mathematically? This is a question from a test paper.

Dr.Peterson

Elite Member
What you just said is perfectly mathematical -- "working" doesn't have to involve symbols.

But you can express this using symbols together with words. Since any square is non-negative (not necessarily positive!), you can say first that $$\displaystyle c^2 \ge 0$$; you still need to show that it is non-zero, which is clear from the initial form they have to find for $$\displaystyle c^2$$, because you are told that a and b are non-zero.

For the other part, you can similarly state that $$\displaystyle \left(\frac{a+b}{a-b}\right)^2 \ge 0$$, and derive the desired inequality from this one.

New member
What you just said is perfectly mathematical -- "working" doesn't have to involve symbols.

But you can express this using symbols together with words. Since any square is non-negative (not necessarily positive!), you can say first that $$\displaystyle c^2 \ge 0$$; you still need to show that it is non-zero, which is clear from the initial form they have to find for $$\displaystyle c^2$$, because you are told that a and b are non-zero.

For the other part, you can similarly state that $$\displaystyle \left(\frac{a+b}{a-b}\right)^2 \ge 0$$, and derive the desired inequality from this one.
Many thanks! Appreciated

Otis

Senior Member
… you can say first that $$\displaystyle c^2 \ge 0$$ …
How can c2 equal zero, when we have c2=-4ab/(a-b)2 with a≠0 and b≠0?

Jomo

Elite Member

If you are subtracting from 1, the value should be less than 1. Also, yes anything square is positive, so the answer can never be less than 0. But does this involve any working? How can I represent my thoughts mathematically? This is a question from a test paper.

You have to be careful with what you say. You made two statements and they are both false.

1) If you are subtracting from 1, the value should be less than 1. Not true at all. 1 - (-5) = 6 and 6 is Not less than 1. Sorry, but it just isn't.

2) yes anything square is positive, so the answer can never be less than 0. I agree that anything squared is never less than 0. However, I do not agree with the statement anything squared is positive. That is not true. 0^2 = 0 and 0 is not positive.

This is mathematics and what you say must be sound. Please be more careful.

Now you are going to argue that in your situation your statements follow. My problem is that you made this statements in general.

For example if had said that if x is not 0 then x^2 is never negative and that x^2 is always positive that would have been fine.

Dr.Peterson

Elite Member
How can c2 equal zero, when we have c2=-4ab/(a-b)2 with a≠0 and b≠0?

You missed my point. I said "you can say first that $$\displaystyle c^2 \ge 0$$; you still need to show that it is non-zero" -- hinting at exactly the reason you give here. There are two parts of this, each with a different reason. The part you mention shows it's non-zero, but is not sufficient to prove it's positive; the mere fact that it is a square proves it is non-negative, but is not sufficient to prove it is non-zero. Together, they prove what we need.

One doesn't have to see an entire proof at once.

Otis

Senior Member
… One doesn't have to see an entire proof at once …
I didn't have to, but that's what happened.

Seriously, though, I did miss your point. I had forgotten the call for a proof, so I was thinking loosely (the fact that a and b must have opposite signs foremost in my mind). Thank you.

$$\;$$