Solutions from lengths of a triangle?

The discriminant of this quadratic is $D=\left(a^2+b^2-c^2\right)^2-4a^2 b^2 = a^4+b^4+c^4-2\left(b^2 c^2+c^2 a^2+a^2 b^2\right)$

If we write $s=\frac12 (a+b+c)$ , then $D=-16s(s-a)(s-b)(s-c)$

which should be familiar as being related to Heron's formula. In fact, $D=-16\Delta^2$

where $\Delta$ is the area of the triangle; so $D<0$ and the equation has no real roots .

Easy way :

A quadratic can have either 2 real solutions, 1 real solution or 0 real solutions

We have three attempts, use them wisely :)

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Better way :

Divide the expression by $2ab$

We get

$\frac{a}{2b} x^2 + \frac{a^2+b^2-c^2}{2ab} x + \frac{b}{2a}=0$

By Cosine Law we have $\cos C = \dfrac{a^2+b^2-c^2}{2ab}$

So the expression turns to

$\frac{a}{2b} x^2 + (\cos C )x + \frac{b}{2a}=0$

Checking the discriminant

$D= \cos^2{C} - 4 \times \frac{a}{2b} \times \frac{b}{2a}$

$D= \cos^2{C} - 1 = - \sin^2{C}$

Therefore $D \leq 0$

So the equation has zero solutions because $\sin C$ cannot equal zero (it equals zero at $C=0,π$ which do not allow a triangle to form)