Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation k^2-2*k+5=0
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Identical expressions
- k^ two - two *k+ five = zero
- k squared minus 2 multiply by k plus 5 equally 0
- k to the power of two minus two multiply by k plus five equally zero
- k2-2*k+5=0
- k²-2*k+5=0
- k to the power of 2-2*k+5=0
- k^2-2k+5=0
- k2-2k+5=0
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Similar expressions
- k^2+2*k+5=0
- k^2-2*k-5=0
k^2-2*k+5=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
This equation is of the form
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
$$a = 1$$
$$b = -2$$
$$c = 5$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$k_{1} = 1 + 2 i$$
Simplify
$$k_{2} = 1 - 2 i$$
Simplify
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
$$a = 1$$
$$b = -2$$
$$c = 5$$
, then
D = b^2 - 4 * a * c =
(-2)^2 - 4 * (1) * (5) = -16
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)
or
$$k_{1} = 1 + 2 i$$
Simplify
$$k_{2} = 1 - 2 i$$
Simplify
Vieta's Theorem
it is reduced quadratic equation
$$k^{2} + k p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = 5$$
Vieta Formulas
$$k_{1} + k_{2} = - p$$
$$k_{1} k_{2} = q$$
$$k_{1} + k_{2} = 2$$
$$k_{1} k_{2} = 5$$
$$k^{2} + k p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = 5$$
Vieta Formulas
$$k_{1} + k_{2} = - p$$
$$k_{1} k_{2} = q$$
$$k_{1} + k_{2} = 2$$
$$k_{1} k_{2} = 5$$
Sum and product of roots
[src]
sum
0 + 1 - 2*I + 1 + 2*I
$$\left(0 + \left(1 - 2 i\right)\right) + \left(1 + 2 i\right)$$
=
2
$$2$$
product
1*(1 - 2*I)*(1 + 2*I)
$$1 \cdot \left(1 - 2 i\right) \left(1 + 2 i\right)$$
=
5
$$5$$
5
The graph