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.

score 2 · Answer 1 · answered Jun 16 '20 at 10:18

2

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

answered Jun 16 '20 at 10:18

José Carlos Santos

427,504

Thank you, but can't any imaginary values of a also satisfy the equation? I want to prove that 'a' can ONLY be real. – Selfish Stoic Jun 16 '20 at 11:39
Sure there are real complex numbers that are roots of the equation: $1\pm4i$. But I assumed, from the title of your question, the you wanted to prove that, if $a\in\Bbb R$, then $a^2-2a+17>0$. Am I wrong? – José Carlos Santos Jun 16 '20 at 11:51
No, my bad, you were right, thank you. – Selfish Stoic Jun 16 '20 at 13:20

score 1 · Answer 2 · answered Jun 16 '20 at 10:21

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!

score 1 · Answer 3 · answered Jun 16 '20 at 10:30

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.

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

3 Answers3