Is it possible to prove $a \in\mathbb R$ in the quadratic?

Question

$a^2−2a+17>0$

I didn't really know how to go about proving this.

At first I tried finding the range of '$a$' but the equation has imaginary roots so that doesn't really help.

Trying to look at it graphically, it's a parabola that doesn't touch the $x$-axis but I just can't prove that '$a$' will never be imaginary for this inequality.

Answer 1

You have $a^2-2a+17=a^2-2a+1+16=(a-1)^2+16>0$. So, the inequality holds for every $a\in\Bbb R$.

Answer 2

Yes, there is! Complete the square: $$(a−1)^2-1+17>0$$ so $$(a−1)^2+16>0$$ This is true for all values of $a$ as any number squared will be greater than or equal to zero. Adding $16$ will obviously ensure the expression is always greater than $0$, so the inequality is true for all real values of $a$. Hope that helped!

Answer 3

On the opposite "the equation has imaginary roots" is quite helpful !

That means that as there are no real roots, the polynomial keeps the same sign for all $a$. Then taking $a=0$, $17>0$ confirms that the sign is positive throughout.