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I'm asked (for homework which isn't graded but instead the basis of a quiz) to directly prove that $2x^2 -4x + 3 > 0$ for all real $x$.

I am VERY new to proofs.

The textbook's only example is a case that was simplified to ( foo )^2 + bar, and it was assumed since ( foo )^2 is always positive that ( foo )^2 must be grater than or equal to bar.

I don't see a way to simplify this into ( something )^2 + something else.

user132181
  • 2,726

3 Answers3

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Complete the square:

$$2x^2-4x+3=(\sqrt 2 x)^2-2\times(\sqrt2 x)\times\frac2{\sqrt2}+\left(\frac2{\sqrt2}\right)^2-\left(\frac2{\sqrt2}\right)^2+3\\=\left(\sqrt 2 x-\left(\frac2{\sqrt2}\right)\right)^2+1$$

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$D = (-4)^2 - 4*3*2 < 0$, hence the polynomial $2x^2 - 4x + 3$ has no real roots, and so its graph does not intersect the horizontal axis. Thus, the value of this polynomial never change its sign, and, for, instance, when $x = 0$ it has the value $3 > 0$, so it always take positive values.

AlexCon
  • 745
3

This is a slight simplification of @Sami's solution:

$$2x^2-4x+3=2(x^2-2x+1)+1=2(x-1)^2+1>0$$

mike
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