Proving that $a^2x^2+(b^2+a^2-c^2)x+b^2=0$ has no real roots if $a+b>c$ and $|a-b|How is this answer derived, and is there any other way to prove it?

Prove that $$a^2x^2+ (b^2 + a^2 - c^2)x + b^2 = 0$$ does not have real roots if $a+b>c$ and $|a-b|<c$.  $a,b,c \in \mathbb{R}$.

Solution I found online:

The discriminant is given as:
$$D = (b^2 + a^2 - c^2)x - 4a^2b^2$$
$$(a^2 + b^2)^2 - 2c^2(a^2 + b^2) + c^4 - 4a^2b^2$$
Let $c^2= t$.
You get : $$t^2 - 2t(a^2 +b^2) + (a^2 + b^2)^2$$ has to be less than zero since $D < 0$ means nonreal roots. Then, you get $$(a-b)^2 < t < (a+b)^2.$$
Hence, you get the condition satisfied.

My questions regarding this proof :
It is right to say that $a,b,c$ have to be some particular numbers only. Only one type. By here, what I mean to say is that let $a=5$, $b =6$ or $c=7$ or another set of numbers , $a=6$ , $b=7$ , $c=4$. Now , according to the condition, it may or may not be possible that one of the conditional value satisfies the condition since $a , b$ and $c$ can have infinite values, right? There can be infinite quadratic equations but not all will satisfy this condition of our question.
If not, can we find out whatever those numbers value can be and which are those numbers which satisfy the condition and how many types of this condition is possible. I think it is impossible since that can be infinite.
I don’t think also that it is true that there will be only two conditions. If I know the values of each variable , I can create a few more conditions. Here , I mean to say that if I have a question with all of its values in numbers except $x$. Like $a, b, c$ has a numerical value. You don’t need any other value. Making condition like $a+b>c$. Who knows ? I can also create another condition like $a*b*c>0$. Right ? Then, our answer will not be the same if this condition was given as an alternative to some other one.
How did you know that in the end you would get this result only ? This answer is like a luck. I am not satisfied with this way of deriving. I am looking for alternate solutions of the above question or id you could help me to understand this proof as a right one.
 A: I'll answer the paragraphs first, then the question.

Paragraphs

It is right ... condition of our question.

Of course, that's correct. Not all $a,b,c$ will satisfy this equation. The reason why the statement is useful is that some $a,b,c$ do satisfy the statement (extreme example : $a = 10^{10} , b = 10^{10}+1  , c=2$)!

If not ... can be infinite.

Yes, of course we can find the numbers satisfying the condition (more on this later, so I'll keep the suspense, Teresa style) and there are, in fact , infinitely many of them. However, it's fair to say that the infiniteness means that you can't take a case-by-case
approach, there are infinitely many cases! So you'll have to contend with the algebraic approach.

I don’t think ... some other one.

Indeed, if the conditions change, then the situation changes as well : so you change your condition to $abc>0$ or something, then you are probably guaranteed some other properties, but probably NOT this condition of having no real roots.

How did ... proof as a right one.

The proof does seem to come a little out of thin air , but some part of it was a little more logical than the rest, namely taking the discriminant. I'll explain that.

Question
Yes, the online approach works, and it's quite simple :

*

*No real roots happens if and only if the discriminant is negative.


*Use all the algebra we know and show that the discriminant is negative.
But how do we justify the algebra? Well, let's look at $a+b>c$ and $|a-b|<c$. The common thing in these conditions is that combining them, $|a-b|<c<a+b$.
So, when we look at the discriminant $(b^2+a^2-c^2)^2 - 4b^2a^2$, the important thing to observe is that isolating $c$ will be giving us the best chance of working out an inequality : separate all the $c$s, substitute the inequalities , and see if things work out.
To do that, obviously one wants to expand : $$
(b^2+a^2-c^2)^2 = (\color{blue}{(b^2+a^2)} - \color{green}{c^2})^2
$$
and then use the formula for two variables. Expanding that, one gets : $$(b^2+a^2)^2 + c^4 - 2c^2(a^2+b^2) - 4a^2b^2$$, and remember we are focusing on $c$ as the variable we feel can be controlled the strongest. It's quite easy to see that the above expression is a quadratic in $c^2$, so if $t = c^2$ then the expression is $$((b^2+a^2)^2 - 4a^2b^2) + t^2 - 2t(a^2+b^2)$$ which becomes, using the identity $(x+y)^2 - 4xy = (x-y)^2$, as $$t^2 - 2t(a^2+b^2) + (a^2-b^2)^2$$
which doesn't match with your attempt, but now there's actually a factorization available using the fact that $2(a^2+b^2) = (a+b)^2 + (a-b)^2$ and difference of squares (that $(a^2-b^2) = (a-b)(a+b)$). More precisely, we have :
\begin{align}
& t^2 - 2t(a^2+b^2) + (a^2-b^2)^2 \\ =& t^2 - t ((a+b)^2 + (a-b)^2) + (a+b)^2(a-b)^2 \\ &= (t - (a+b)^2) (t - (a-b)^2)
\end{align}
Now we have the product of two terms, and NOW it's clear why we have the conditions at the start of the question : $a+b>c$ implies that $(t- (a+b)^2)<0$, and $|b-a|<c$ implies that $(t - (a-b)^2)> 0$. So one of the terms in the product is positive, the other is negative. So the discriminant is negative, and we are done.

Further introspection
Further introspection, however, is necessary for a couple of reasons. The method above is nice, but not always the most efficient to think about.
An alternate method comes when you think of geometry.
NOTE : This method is more intuitive and involves simpler computation than the other method , but it's slightly longer and goes through a lot of little steps which are by themselves VERY interesting.

Claim : The set of all $a,b,c$ such that $|a-b|<c$ and $a+b>c$ is precisely the set of  of all positive real triples, for which one can form a triangle with those lengths.

Proof : Suppose $|a-b|<c$ and $a+b>c$. Then $c>0$ is clear, since $c>|a-b| \geq 0$. From $a+b>c$ we know at least one of $a,b >0$ must be true. Let's say $b>0$. Then, $a\leq 0$ cannot happen : put things on a number line and see why $|b-a|<c$ is violated if $a+b>c$ but $a \leq 0$. So , in fact $a > 0$ must happen as well. You can do something similar if $a>0$ instead of $b$.
So, $a,b,c$ are all positive. $a+b>c$ is true, and from $|a-b|<c$ we know from the triangle inequality that $a<b+c$ and $b<a+c$ as well. These are precisely the conditions for triangle formation. The converse is easily seen to be true. $\blacksquare$
So, we have a geometric interpretation for our domain of interest. Basically, if $a,b,c$ satisfy the conditions, then we can imagine a triangle with side lengths $a,b,c$, and with all angles between $0$ and $180$ degrees (so no obsolete triangles). Let's call the angles opposite $a,b,c$ as $A,B,C$ respectively . What about our equation looks geometric?
Recall the cosine rule : $\cos C  = \frac{b^2+a^2 - c^2}{2ab}$. This very useful looking equation immediately suggests to us that we should look to geometrically interpret the equation by performing this substitution :
$$
a^2x^2 + (b^2+a^2-c^2)x + b^2 = 0 \to a^2x^2 + 2ab\cos C x + b^2 = 0
$$
we write this to make things more illustrative , as :
$$
(ax)^2 + 2 \cos C (ax)b + b^2 = 0
$$
(NOTE : If you want to see an easier proof, then see the end of the answer, but what follows is my first proof)
and now we perform the completion of square procedure, which basically realizes that you have $(ax)^2$ and $b^2$ so all you need to do is subtract and add $2axb$. Once you do that, you get :
$$
(ax)^2 + b^2 + 2 \cos C(ax)b = [(ax)^2 + 2axb +  b^2] + [-2axb + 2\cos C(ax)b] = 0
$$
both $[\cdot]$ brackets factorize, giving :
$$
(ax+b)^2 + 2axb(-1+\cos C) = 0 \to (ax+b)^2 = (2ab(1-\cos C))x 
$$
following transposition. I claim this cannot have a real solution. Let's see why. First recognize that $1-\cos C \geq 0$ for all values of $C$.
Suppose that $x < 0$, the easy case. Then, we know that $2ab(1-\cos C)x \leq 0$, while $(ax+b)^2 \geq 0$. So the only way that equality can happen is if $(ax+b)^2 = 0 = 2ab(1-\cos C)x$. Which implies that $1-\cos C = 0$ , but then $\cos C = 1$ implies that the angle at $C$ is 0 degrees, which is a contradiction.
Suppose that $x > 0$. Then note that $(ax-b)^2 \geq 0$ so rearranging, $(ax)^2+b^2 \geq 2abx \geq 2ab\cos C x$ because $x > 0$ and $1 \geq \cos C$. If equality occurs, then we must have $\cos C = 1$ so that angle $C = 0$, again a contradiction.
Clearly $x=0$ is not a root. Hence we are done.

To summarize : the fact that $a,b,c$ form a triangle restricts $b^2+a^2-c^2$ in a fashion we can see geometrically, since $\cos C$ expresses the relation between this quantity and $a$ and $b$.
Thus, we used :

*

*Geometric interpretation of the domain


*The cosine rule


*Completion of squares, and the :


*Creation of cases by splitting the sign of $x$
to solve this problem without using any kind of supremely complicated mathematics : no discriminant, no awkward multiplications, and most importantly, a proof that went through a completely different domain of mathematics.
I'm not sure that in a competitive exam, one can chalk out an approach like the geometric one I mention (which, extremely unfortunately, means that the online approach is in fact the correct approach to be taken, and one needs to be good at algebra for seeing how it works). Therefore, my approach is being placed only to intuit the source of the question from a geometric point of view, and solve it from the same point of view. I hope this was helpful as an alternate view of the same situation.

EDIT : From the hint given by dxiv below (which I thank them for very much!) the question becomes much, much easier. The hint asks us to complete the square, but using a different pair of terms. Indeed,  we try to complete the square , but not using $(ax)^2$ and $b^2$, but instead, $(ax)^2$ and $2abx\cos C$ :
\begin{align}
(ax)^2 + 2ab(\cos C)x + b^2 & = [(ax)^2 + 2ab(\cos C)x + b^2(\cos^2 C)] + [b^2 - b^2\cos^2 C] \\
& = [(ax + b \cos C)^2] + [b^2(1-\cos^2 C)] \\
& = \color{blue}{(ax+b \cos C)^2 + (b \sin C)^2}  
\end{align}
is a sum of squares! That's remarkable, because a sum of squares is always non-negative.
In particular, if $(ax+b\cos C)^2 + (b \sin C)^2 =0$ then we MUST have that $ax+b \cos C = 0$ and $b \sin C = 0$. But the second part implies that $\sin C = 0$ since $b>0$, so $C = 0$, a contradiction!
Therefore, the equation can have no real root. So, you can see that trigonometric manipulation actually gets us much, much farther than in my first attempt.
A: The line of proof works, though the proof has a typo as quoted, and it may not be written in the easiest way to follow.

The discriminant is given as: $D = (b^2 + a^2 - c^2)x - 4a^2b^2$

Here is the typo, and the discriminant can not depend on $x$.
The expression above is wrong (though what's used afterwards appears to be based on the correct expression). This was supposed to be $D = (b^2 + a^2 - c^2)^2 - 4a^2b^2$.

Prove that $a^2x^2+ (b^2 + a^2 - c^2)x + b^2 = 0$ does not have real roots if $a+b>c$ and $|a-b|<c$.  $a,b,c \in \mathbb{R}$.

What is to be proved is the one-way implication: if $\;a+b>c\;$ and $\;\mid a-b\mid<c\;$ then the equation $\;a^2x^2+ (b^2 + a^2 - c^2)x + b^2 = 0\;$ has no real roots.
Without repeating the calculations in detail, the proof goes through the following steps.

*

*The absolute value of a number can never be negative, so $\;0 \le \mid a-b\mid<c \implies$ $\require{bbox}\bbox[5px, border: 1px solid black]{c \gt 0}\,$.


*Inequalities between positive numbers can be multiplied, so $\;\mid a-b\mid<c$ $\implies \mid a-b\mid^2<c^2$ $\implies \bbox[5px, border: 1px solid black]{(a-b)^2 \lt c^2}\,$, since the square of a real number is equal to the square of its absolute value.


*Given $a+b\gt c$ and since $c \gt 0$ the inequality can be squared, so $\;\bbox[5px, border: 1px solid black]{(a+b)^2 \gt c^2}\;$.


*The discriminant is the difference of two squares, which can be factored as $D=\left(\left(a+b\right)^2-c^2\right)\left(\left(a-b\right)^2-c^2\right)\,$. It follows from the previous inequalities that the first factor is positive and the second one is negative, so $\;\bbox[5px, border: 1px solid black]{D \lt 0}\;$.
A quadratic with negative discriminant does not have real roots, which concludes the proof.
