Show that $3x^2-4x+2$ is always greater than $0$.

Question

How do I show that the function $3x^2-4x+2$ is always greater than $0$?

Answer 1

Let $f(x)=3x^2-4x+2$. Now, since this is a quadratic function, so it should have only one maxima or minima. So, we differentiate the function to find out what is the point that reaches the maxima or minima and also, find out whether it is a maximum of minimum value to reach the solution.

$f'(x)=6x-4$
So, when $f'(x)=0$, we have a maxima or a minima there. The point is $x=\dfrac23$
$f''(x)=6$. Now, $f''(x)$ is a positive quantity, so the point is $f''(\dfrac23)>0$, so the function reaches minima at $x=\dfrac23$.

At $x=\dfrac23,f(x)=3\bigg(\dfrac23\bigg)^2-4\bigg(\dfrac23\bigg)+2>0$.

So, as desired, the function is always greater than $0$.

Answer 2

It is not. $x = 1 \implies 3x^2-4x+2 = 3-4+2 = 1$.

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In fact, there are exactly two values of $x$ for which this function is zero, and both those values are complex, meaning that there are no real values for which the function is zero.

Since there are no real roots of the function, and the function is greater than zero at some values (e.g. $x=1$), and the function is continuous, then it must always be positive.

I will leave it to you to use the quadratic formula (hint!) to show that there are no real roots.

Answer 3

Hint: Complete the square. We have that

$$3x^2 - 4x + 2 = 3(x-2/3)^2 + 2/3.$$

Recall that if $u \in \mathbb{R}$, then $u^2 \geq 0$. So for every $x \in \mathbb{R}$,

$$3x^2 - 4x + 2 = 3(x-2/3)^2 + 2/3 \geq 3 \cdot 0 + 2/3 = 2/3 >0.$$

Answer 4

$$3x^2-4x+2=\left(\sqrt{3}x+\frac{2}{\sqrt{3}}\right)^2+\frac{2}{3}$$

Answer 5

Maybe just another "redundant" method can be this.

Calculate the discriminant of the polynomial $(b^2-4ac=-12)$. Since it is negative then the polynomial doesn't have any real roots. So this means the polynomial never touch the x-axis. So it is always above or under this axis. Calculate the value of the polynomial in one point (1 for example) to conclude.

two little words about the reasoning here the main point in this reasoning is the fact that polynomials functions are continuous from $\mathbb{R}$ to $\mathbb{R}$. (Even in more general context but this is not the case). So by the intermediate value theorem you can conclude that the function is always greater than zero. Even if you don't know anything about topology it is a good thing "learning" this theorem, which gives an intuition about "what is the meaning of being continuous in $\mathbb{R}$" (clearly it is not a definition neither a sufficient condition, but it is necessary)

NB I'm assuming you know what the (sign of the ) discriminant of a polynomial function means :)