# Find all numbers $x$ for which $x^2 - 2x + 2 >0$

Find all numbers $x$ for which $x^2 - 2x + 2 >0$.

Usually I can solve this type of questions simply by finding the zero's through factoring/quadratic equation and then imagining the graph. But what do I do if the discriminant ($b^2 - 4ac$) is negative so I get a negative number under the square root sign in the quadratic formula?

Thanks a lot

• Do you know what the sign of the discriminant tells you about the number of roots the equation has? – jacer21 Nov 29 '16 at 21:09
• if the b^2-4ac is negative then the function f(x) does not across the 0 line so the f(x) should either all above 0 or all below 0 – displayname Nov 29 '16 at 21:12
• Have you tried graphing the function? – Hugh Nov 29 '16 at 21:13
• Thanks to everybody! jacer21 completely forgot about that. And graphing the function indeed showed it clearly – stef Nov 29 '16 at 21:37
• If the discriment is negative then the value is never 0.... which means the function is always positive or always negative. – fleablood Nov 29 '16 at 21:40

You can complete the square, which changes the inequality to $$(x-1)^2+1>0$$ Note that this is true for all $x$, because $(x-1)^2\ge 0$.

The discriminant of $x^2-2x+2=0$ is $b^2-4ac=(-2)^2-4(2)(1)=-4$. Meaning this quadratic has no solution.

And since $a>0$, this parabola opens upwards. Meaning any $x$ input produces a $y$ value greater than $0$.

Let $f(x) = ax^2 + bx +c$. If $f(x)$ never equals $0$ then the graph of this parabola can never cross the $x$ axis. So $f(x)$ is either always positive or it is always negative.

$f(x) = x^2 -2x + 2$ is never equal to zero as $b^2 - 4ac = 4 - 8 = -4 < 0$ as you pointed out. So it is either always positive or always negative.

$f(0) = 2 > 0$ so $f(x)$ is always positive.

So $x^2 -2x + 2 > 0$ for all real $x$.