How do you solve $x^{2} + 3 > 2 x$ $x^2+3>2x$ using a sign chart?

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Answer 1 · 2017-02-27T10:05:05

The solution is $x in RR$

Let's rewrite the equation

$x^2-2x+3>0$

Let $f(x)=x^2-2x+3$

We need the roots of

$x^2-2x+3=0$

Let's calculate the discriminant

$Delta=b^2-4ac=(-2)^2-4(3)(1)$

$=4-12=-8$

As, $Delta<0$ , there are no real roots

So,

$AA x in RR$ , $f(x)>0$

graph{x^2-2x+3 [-6.176, 6.31, -0.31, 5.93]}

How do you solve x^2+3>2xx2+3>2xx^2+3>2x using a sign chart?