3x^2-4x+1 equation
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The solution
Detail solution
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 3$$
$$b = -4$$
$$c = 1$$
, then
D = b^2 - 4 * a * c =
(-4)^2 - 4 * (3) * (1) = 4
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 1$$
Simplify$$x_{2} = \frac{1}{3}$$
Simplify
Vieta's Theorem
rewrite the equation
$$3 x^{2} - 4 x + 1 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - \frac{4 x}{3} + \frac{1}{3} = 0$$
$$p x + x^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{4}{3}$$
$$q = \frac{c}{a}$$
$$q = \frac{1}{3}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = \frac{4}{3}$$
$$x_{1} x_{2} = \frac{1}{3}$$
$$x_{1} = \frac{1}{3}$$
$$x_{2} = 1$$
Sum and product of roots
[src]
$$\left(0 + \frac{1}{3}\right) + 1$$
$$\frac{4}{3}$$
$$1 \cdot \frac{1}{3} \cdot 1$$
$$\frac{1}{3}$$