Prove that if $a$,$b$ and $c$ are consecutive positive integers then the roots of the quadratic equation of the form $ax^2 + bx + c = 0$ has Complex roots
My attempt:
In my attempt to grapple with this intriguing problem, I embarked on a journey through the realms of quadratic equations and consecutive integers. At first glance, the task seemed daunting, but I was determined to unravel its mysteries.
The discriminant of a quadratic equation determines the nature of its roots: positive for two real roots, zero for repeated roots, and negative for complex roots. So in this question I had to explore the scenario where the roots are complex.
First, I considered the relationship between the coefficients $a$, $b$, and $c$. Since they are consecutive positive integers, we can express them as (a), (a + 1), and (a + 2), respectively. Substituting these values into the discriminant formula, I obtained:
$D = (a + 1)^2 - 4a(a + 2)$
Expanding and simplifying this expression, I arrived at: $D = a^2 + 2a + 1 - 4a^2 - 8a$ $D = -3a^2 - 6a + 1$
Despite my efforts, I found myself faced with a formidable challenge. The expression $D = -3a^2 - 6a + 1$ appeared enigmatic, defying easy interpretation. However, I refused to be deterred. I embarked on a meticulous analysis, exploring various avenues in search of insights but I failed.
You were almost there!
when $a,b,c$ are consecutive possitive integers the trinomial can be written
$nx^2+(n+1)x+(n+2)=0 $ ,
The discriminant in this case would be $D=(n+1)^2-4n(n+2)=-3n^2-6n+1 $
We know that $n$ is a possitive integer so $-3n^2,-6n+1 < 0 \implies D<0 $ We can also say that when $n\ge 1 , D \le -8<0$
So indeed the trinomial has $2$ complex roots
If you are still not convinced
We can let $n+1=y$ or $n+2=y$ and the discriminant is still negative