If $c$ is positive and $2ax^2 + 3bx + 5c$ does not have any real roots, then prove that $2a - 3b + 5c > 0$.
2026-04-06 12:42:58.1775479378
Quadratic Equation - Prove the following
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If $c$ is positive, then you get that the polynomial $2ax^2+3bx+5c$ has a stricly positive value for $x=0$.
Since it is continuous and does not have real roots, you know that it will always be strictly positive (for all $x$).
You can then take $x=-1$, and you get
$$2a-3b+5c>0.$$