I'm not much into Algebra and Functions so I got confused in this one. The question ( I'll write it exactly how it asks ) says,
The function f is defined by $f : x \to 2x^2 -6x + 5$ for $x \in \mathbb{R}$. Find the set of values of P for which the equation $f(x) = p$ has no real roots.
Please provide answers in simple forms. And Please provide it as soon as you can. I have exams after a few days and I don't have concept on this one.
Thanks in Advance!
Best Answer
the equation $f(x)=2x^2 -6x + 5=p$ leads to a second order equation $2x^2 -6x + 5-p=0$. This equation has no real solutions when the discriminant is smaller than zero (if $ax^2+bx+c=0$, the discriminant is $b^2-4ac$).