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k^2-4*k+3=0

k^2-4*k+3=0 equation

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Numerical solution:

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The solution

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k  - 4*k + 3 = 0
$$k^{2} - 4 k + 3 = 0$$
Detail solution
This equation is of the form
a*k^2 + b*k + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$k_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$k_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -4$$
$$c = 3$$
, then
D = b^2 - 4 * a * c = 

(-4)^2 - 4 * (1) * (3) = 4

Because D > 0, then the equation has two roots.
k1 = (-b + sqrt(D)) / (2*a)

k2 = (-b - sqrt(D)) / (2*a)

or
$$k_{1} = 3$$
Simplify
$$k_{2} = 1$$
Simplify
Vieta's Theorem
it is reduced quadratic equation
$$k^{2} + k p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = -4$$
$$q = \frac{c}{a}$$
$$q = 3$$
Vieta Formulas
$$k_{1} + k_{2} = - p$$
$$k_{1} k_{2} = q$$
$$k_{1} + k_{2} = 4$$
$$k_{1} k_{2} = 3$$
The graph
Rapid solution [src]
k1 = 1
$$k_{1} = 1$$
k2 = 3
$$k_{2} = 3$$
Sum and product of roots [src]
sum
0 + 1 + 3
$$\left(0 + 1\right) + 3$$
=
4
$$4$$
product
1*1*3
$$1 \cdot 1 \cdot 3$$
=
3
$$3$$
3
Numerical answer [src]
k1 = 3.0
k2 = 1.0
k2 = 1.0
The graph
k^2-4*k+3=0 equation