quadratics equation tricky problem

Question

I am confused with this question- if $ax^2+bx+c$ have no real roots then-

$1+c/a+b/a$ is--

a. Positive

b. Negative

c. Zero

d. Can.t say

I tried attempting it as follows $b^2-4ac<0$

so $(b/2a)^2<c/a$

so $c/a$ is Positive. But what about $b/a$?

Answer 1

Hint. If $ax^2+bx+c=0$, has no real roots, i.e., $b^2-4ac<0$, then $$ x^2+\frac{b}{a}x+\frac{c}{a}=\left(x-\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}>0, $$ for all $x\in\mathbb R$.

Answer 2

Hint: For very large positive or negative $x$ note that the quadratic has a value which is the same sign as $a$. Now consider also the value when $x=1$.

Or divide through by $a$ first.

Answer 3

If the polynomial (call it $P(x)$) has no real roots, then its values have the same sign as $a$.

But $$1+\frac ba+\frac ca=\frac1aP(1)>0$$