Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation x^3+9*x^2+27*x+27=0
-
Equation x^2-7*x+6=0
-
Equation x^2+3x+1=0
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Equation x^3+x-10=0
- Express {x} in terms of y where:
- -11*x+16*y=-5
- -12*x-9*y=-3
- -5*x+15*y=-15
- -12*x+2*y=14
- Factor polynomial:
- x^3+x-10
-
Identical expressions
- x^ three +x- ten = zero
- x cubed plus x minus 10 equally 0
- x to the power of three plus x minus ten equally zero
- x3+x-10=0
- x³+x-10=0
- x to the power of 3+x-10=0
- x^3+x-10=O
-
Similar expressions
- x^3+x+10=0
- x^3-x-10=0
x^3+x-10=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\left(x^{3} + x\right) - 10 = 0$$
transform
$$\left(x + \left(x^{3} - 8\right)\right) - 2 = 0$$
or
$$\left(x + \left(x^{3} - 2^{3}\right)\right) - 2 = 0$$
$$\left(x - 2\right) + \left(x^{3} - 2^{3}\right) = 0$$
$$\left(x - 2\right) \left(\left(x^{2} + 2 x\right) + 2^{2}\right) + \left(x - 2\right) = 0$$
Take common factor -2 + x from the equation
we get:
$$\left(x - 2\right) \left(\left(\left(x^{2} + 2 x\right) + 2^{2}\right) + 1\right) = 0$$
or
$$\left(x - 2\right) \left(x^{2} + 2 x + 5\right) = 0$$
then:
$$x_{1} = 2$$
and also
we get the equation
$$x^{2} + 2 x + 5 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
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
$$x_{2} = -1 + 2 i$$
$$x_{3} = -1 - 2 i$$
The final answer for x^3 + x - 10 = 0:
$$x_{1} = 2$$
$$x_{2} = -1 + 2 i$$
$$x_{3} = -1 - 2 i$$
$$\left(x^{3} + x\right) - 10 = 0$$
transform
$$\left(x + \left(x^{3} - 8\right)\right) - 2 = 0$$
or
$$\left(x + \left(x^{3} - 2^{3}\right)\right) - 2 = 0$$
$$\left(x - 2\right) + \left(x^{3} - 2^{3}\right) = 0$$
$$\left(x - 2\right) \left(\left(x^{2} + 2 x\right) + 2^{2}\right) + \left(x - 2\right) = 0$$
Take common factor -2 + x from the equation
we get:
$$\left(x - 2\right) \left(\left(\left(x^{2} + 2 x\right) + 2^{2}\right) + 1\right) = 0$$
or
$$\left(x - 2\right) \left(x^{2} + 2 x + 5\right) = 0$$
then:
$$x_{1} = 2$$
and also
we get the equation
$$x^{2} + 2 x + 5 = 0$$
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
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.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = -1 + 2 i$$
$$x_{3} = -1 - 2 i$$
The final answer for x^3 + x - 10 = 0:
$$x_{1} = 2$$
$$x_{2} = -1 + 2 i$$
$$x_{3} = -1 - 2 i$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 1$$
$$v = \frac{d}{a}$$
$$v = -10$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 1$$
$$x_{1} x_{2} x_{3} = -10$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 1$$
$$v = \frac{d}{a}$$
$$v = -10$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 1$$
$$x_{1} x_{2} x_{3} = -10$$
Rapid solution
[src]
x1 = 2
$$x_{1} = 2$$
x2 = -1 - 2*I
$$x_{2} = -1 - 2 i$$
x3 = -1 + 2*I
$$x_{3} = -1 + 2 i$$
x3 = -1 + 2*i
Sum and product of roots
[src]
sum
2 + -1 - 2*I + -1 + 2*I
$$\left(2 + \left(-1 - 2 i\right)\right) + \left(-1 + 2 i\right)$$
=
0
$$0$$
product
2*(-1 - 2*I)*(-1 + 2*I)
$$2 \left(-1 - 2 i\right) \left(-1 + 2 i\right)$$
=
10
$$10$$
10
The graph