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 [MATH]c^2 = 1 - \left(\frac{a+b}{a-b}\right)^2[/MATH], can [MATH]c^2[/MATH] ever be greater than 1? That part should be easy. (By the way, they told you that [MATH]c \ne 1[/MATH], so the "inclusive" part was not really necessary.)
And it should be even easier to see that [MATH]c^2 > 0[/MATH].
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.
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 [MATH]c^2 \ge 0[/MATH]; you still need to show that it is non-zero, which is clear from the initial form they have to find for [MATH]c^2[/MATH], because you are told that a and b are non-zero.
For the other part, you can similarly state that [MATH]\left(\frac{a+b}{a-b}\right)^2 \ge 0[/MATH], and derive the desired inequality from this one.
How can c2 equal zero, when we have c2=-4ab/(a-b)2 with a≠0 and b≠0?… you can say first that [MATH]c^2 \ge 0[/MATH] …
You have to be careful with what you say. You made two statements and they are both false.Many thanks for your reply.
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.
Thanks in advance!
You missed my point. I said "you can say first that [MATH]c^2 \ge 0[/MATH]; 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.How can c2 equal zero, when we have c2=-4ab/(a-b)2 with a≠0 and b≠0?
I didn't have to, but that's what happened.… One doesn't have to see an entire proof at once …