Mister Exam

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k^2-4*k+13=0 equation

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You have entered [src]

 2               
k  - 4*k + 13 = 0

k^{2} - 4 k + 13 = 0

Simplify

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 = 13

, then

D = b^2 - 4 * a * c =

(-4)^2 - 4 * (1) * (13) = -36

Because D<0, then the equation
has no real roots,
but complex roots is exists.

k1 = (-b + sqrt(D)) / (2*a)

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

k_{1} = 2 + 3 i

k_{2} = 2 - 3 i

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 = 13

Vieta Formulas

k_{1} + k_{2} = - p

k_{1} k_{2} = q

k_{1} + k_{2} = 4

k_{1} k_{2} = 13

The graph

Plot the graph f1(x)

Plot the graph f2(x)

Rapid solution [src]

k1 = 2 - 3*I

k_{1} = 2 - 3 i

Simplify

k2 = 2 + 3*I

k_{2} = 2 + 3 i

Simplify

Sum and product of roots [src]

sum

0 + 2 - 3*I + 2 + 3*I

\left(0 + \left(2 - 3 i\right)\right) + \left(2 + 3 i\right)

4

product

1*(2 - 3*I)*(2 + 3*I)

1 \cdot \left(2 - 3 i\right) \left(2 + 3 i\right)

13

Numerical answer [src]

k1 = 2.0 + 3.0*i

k2 = 2.0 - 3.0*i

k2 = 2.0 - 3.0*i

The graph