Problem: If $ax^2+bx+c=0$ and $2x^2 +3x+4=0$ have a common root where $a,b,c \in \Bbb N$, find least value of $a+b+c$
Solution: Here $2x^2 +3x+4=0$ will give complex roots
These roots will be in pair
Both equations have a common root
$ax^2+bx+c=0$ also have these complex pairs
This means least value will be obtained at $a=2$, $b=3$ and $c=4$
Least value of $a+b+c= 9$
Am I doing right?
On
Big gap! You implicitly assume that both equations will share two roots. This is the crux of the matter and it requires rigorous proof! (the accepted answer has the same gap). Here is one way.
Hint $ $ Call them $\,f(x)\,$ and $\,g(x).\,$ A common root is a also a common root of the polynomial $\,h\,$ obtained by eliminating their quadratic terms: $\,h = f-(a/2) g.\,$ Since the discriminant of $\,g\,$ is negative, both roots of $g$ are non-rational. So the only way either can be a root of the polynomial $\,h\,$ of degree $\le 1\,$ is if $\,h = 0.\,$ Thus $\,f = (a/2) g\,$ is a constant multiple of $\,g.\,$ The rest is straightforward.
Remark $\ $ More generally any common root of $\,f,g\,$ is also a root of every one of their linear combinations $\, h_1 f + h_2 g.\,$ But, by Bezout, we know that $\,\gcd(f,g)\,$ has that form, hence $\,f(a) = 0 = g(a)\,\Rightarrow\, \gcd(f,g)(a) = 0.$
What about actually finding the roots? After all this is high school stuff:
$$2x^2+3x+4=0\iff x_{1,2}=\frac{-3\pm\sqrt{23}\,i}4$$
and thus
$$ax^2+bx+c=a(x-x_1)(x-x_2)$$
By Vieta's formulas:
$$b=-(x_1+x_2)a=\frac32a\implies a\;\;\text{is even positive}$$
$$c=ax_1x_2=2a\implies c\;\;\text{is at least}\;\;4\;,\;\text{by the above line}$$
Well, there you go...