Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 9,4x-7,8x+0,52=1
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Equation k^2+4=0
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Equation k^2+9=0
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Equation sinx=-1/2
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Identical expressions
- k^ two + nine = zero
- k squared plus 9 equally 0
- k to the power of two plus nine equally zero
- k2+9=0
- k²+9=0
- k to the power of 2+9=0
- k^2+9=O
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Similar expressions
- k^2-9=0
k^2+9=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
$$a = 1$$
$$b = 0$$
$$c = 9$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$k_{1} = 3 i$$
Simplify
$$k_{2} = - 3 i$$
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
$$a = 1$$
$$b = 0$$
$$c = 9$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (1) * (9) = -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)
or
$$k_{1} = 3 i$$
Simplify
$$k_{2} = - 3 i$$
Simplify
Vieta's Theorem
it is reduced quadratic equation
$$k^{2} + k p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 9$$
Vieta Formulas
$$k_{1} + k_{2} = - p$$
$$k_{1} k_{2} = q$$
$$k_{1} + k_{2} = 0$$
$$k_{1} k_{2} = 9$$
$$k^{2} + k p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 9$$
Vieta Formulas
$$k_{1} + k_{2} = - p$$
$$k_{1} k_{2} = q$$
$$k_{1} + k_{2} = 0$$
$$k_{1} k_{2} = 9$$
Sum and product of roots
[src]
sum
0 - 3*I + 3*I
$$\left(0 - 3 i\right) + 3 i$$
=
0
$$0$$
product
1*-3*I*3*I
$$3 i 1 \left(- 3 i\right)$$
=
9
$$9$$
9
The graph