Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x^2=6
-
Equation -9(8-9x)=4x+5
-
Equation 9x-6(x-1)=5(x+2)
-
Equation k^2-2*k+2=0
- Express {x} in terms of y where:
- -10*x+18*y=6
- 9*x-18*y=6
- -1*x+19*y=-10
- 4*x+13*y=13
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Identical expressions
- k^ two - two *k+ two = zero
- k squared minus 2 multiply by k plus 2 equally 0
- k to the power of two minus two multiply by k plus two equally zero
- k2-2*k+2=0
- k²-2*k+2=0
- k to the power of 2-2*k+2=0
- k^2-2k+2=0
- k2-2k+2=0
- k^2-2*k+2=O
-
Similar expressions
- k^2-2*k-2=0
- k^2+2*k+2=0
k^2-2*k+2=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
$$a = 1$$
$$b = -2$$
$$c = 2$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$k_{1} = 1 + i$$
$$k_{2} = 1 - i$$
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
$$a = 1$$
$$b = -2$$
$$c = 2$$
, then
D = b^2 - 4 * a * c =
(-2)^2 - 4 * (1) * (2) = -4
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
$$k_{1} = 1 + i$$
$$k_{2} = 1 - i$$
Vieta's Theorem
it is reduced quadratic equation
$$k^{2} + k p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = 2$$
Vieta Formulas
$$k_{1} + k_{2} = - p$$
$$k_{1} k_{2} = q$$
$$k_{1} + k_{2} = 2$$
$$k_{1} k_{2} = 2$$
$$k^{2} + k p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = 2$$
Vieta Formulas
$$k_{1} + k_{2} = - p$$
$$k_{1} k_{2} = q$$
$$k_{1} + k_{2} = 2$$
$$k_{1} k_{2} = 2$$
Sum and product of roots
[src]
sum
1 - I + 1 + I
$$\left(1 - i\right) + \left(1 + i\right)$$
=
2
$$2$$
product
(1 - I)*(1 + I)
$$\left(1 - i\right) \left(1 + i\right)$$
=
2
$$2$$
2
The graph