Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation x^2-7*x+10=0
-
Equation 5x^2-8x+3=0
-
Equation 3/5*y=29/10-1/5
-
Equation -12x+11=-(2-x)
- Express {x} in terms of y where:
- 12*x-20*y=9
- -3*x-10*y=1
- -11*x+16*y=3
- 11*x-6*y=5
-
Identical expressions
- k^ two -2k+ six = zero
- k squared minus 2k plus 6 equally 0
- k to the power of two minus 2k plus six equally zero
- k2-2k+6=0
- k²-2k+6=0
- k to the power of 2-2k+6=0
- k^2-2k+6=O
-
Similar expressions
- k^2+2k+6=0
- k^2-2k-6=0
k^2-2k+6=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
$$a = 1$$
$$b = -2$$
$$c = 6$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$k_{1} = 1 + \sqrt{5} i$$
$$k_{2} = 1 - \sqrt{5} i$$
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
$$a = 1$$
$$b = -2$$
$$c = 6$$
, then
D = b^2 - 4 * a * c =
(-2)^2 - 4 * (1) * (6) = -20
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
$$k_{1} = 1 + \sqrt{5} i$$
$$k_{2} = 1 - \sqrt{5} i$$
Vieta's Theorem
it is reduced quadratic equation
$$k^{2} + k p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = 6$$
Vieta Formulas
$$k_{1} + k_{2} = - p$$
$$k_{1} k_{2} = q$$
$$k_{1} + k_{2} = 2$$
$$k_{1} k_{2} = 6$$
$$k^{2} + k p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = 6$$
Vieta Formulas
$$k_{1} + k_{2} = - p$$
$$k_{1} k_{2} = q$$
$$k_{1} + k_{2} = 2$$
$$k_{1} k_{2} = 6$$
Sum and product of roots
[src]
sum
___ ___ 1 - I*\/ 5 + 1 + I*\/ 5
$$\left(1 - \sqrt{5} i\right) + \left(1 + \sqrt{5} i\right)$$
=
2
$$2$$
product
/ ___\ / ___\ \1 - I*\/ 5 /*\1 + I*\/ 5 /
$$\left(1 - \sqrt{5} i\right) \left(1 + \sqrt{5} i\right)$$
=
6
$$6$$
6
Rapid solution
[src]
___ k1 = 1 - I*\/ 5
$$k_{1} = 1 - \sqrt{5} i$$
___ k2 = 1 + I*\/ 5
$$k_{2} = 1 + \sqrt{5} i$$
k2 = 1 + sqrt(5)*i
Numerical answer
[src]
k1 = 1.0 + 2.23606797749979*i
k2 = 1.0 - 2.23606797749979*i
k2 = 1.0 - 2.23606797749979*i