k^2-4*k+3=0 equation
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The solution
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 = D − b 2 a k_{1} = \frac{\sqrt{D} - b}{2 a} k 1 = 2 a D − b k 2 = − D − b 2 a k_{2} = \frac{- \sqrt{D} - b}{2 a} k 2 = 2 a − D − b where D = b^2 - 4*a*c - it is the discriminant.
Because
a = 1 a = 1 a = 1 b = − 4 b = -4 b = − 4 c = 3 c = 3 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 k_{1} = 3 k 1 = 3 Simplify k 2 = 1 k_{2} = 1 k 2 = 1 Simplify
Vieta's Theorem
it is reduced quadratic equation
k 2 + k p + q = 0 k^{2} + k p + q = 0 k 2 + k p + q = 0 where
p = b a p = \frac{b}{a} p = a b p = − 4 p = -4 p = − 4 q = c a q = \frac{c}{a} q = a c q = 3 q = 3 q = 3 Vieta Formulas
k 1 + k 2 = − p k_{1} + k_{2} = - p k 1 + k 2 = − p k 1 k 2 = q k_{1} k_{2} = q k 1 k 2 = q k 1 + k 2 = 4 k_{1} + k_{2} = 4 k 1 + k 2 = 4 k 1 k 2 = 3 k_{1} k_{2} = 3 k 1 k 2 = 3
Sum and product of roots
[src]
( 0 + 1 ) + 3 \left(0 + 1\right) + 3 ( 0 + 1 ) + 3
1 ⋅ 1 ⋅ 3 1 \cdot 1 \cdot 3 1 ⋅ 1 ⋅ 3