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How to use it?
- Equation:
-
Equation -15x=45
-
Equation x^2=15
-
Equation x^4=16
-
Equation 3/5*x=1
- Express {x} in terms of y where:
- 18*x+17*y=-14
- -1*x-6*y=18
- 16*x-17*y=-15
- -4*x-4*y=19
-
Identical expressions
- x^ three - six *x^ two - fifteen *x+ one hundred = zero
- x cubed minus 6 multiply by x squared minus 15 multiply by x plus 100 equally 0
- x to the power of three minus six multiply by x to the power of two minus fifteen multiply by x plus one hundred equally zero
- x3-6*x2-15*x+100=0
- x³-6*x²-15*x+100=0
- x to the power of 3-6*x to the power of 2-15*x+100=0
- x^3-6x^2-15x+100=0
- x3-6x2-15x+100=0
- x^3-6*x^2-15*x+100=O
-
Similar expressions
- x^3+6*x^2-15*x+100=0
- x^3-6*x^2-15*x-100=0
- x^3-6*x^2+15*x+100=0
x^3-6*x^2-15*x+100=0 equation
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The solution
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3 2 x - 6*x - 15*x + 100 = 0
$$\left(- 15 x + \left(x^{3} - 6 x^{2}\right)\right) + 100 = 0$$
Detail solution
Given the equation:
$$\left(- 15 x + \left(x^{3} - 6 x^{2}\right)\right) + 100 = 0$$
transform
$$\left(- 15 x + \left(\left(- 6 x^{2} + \left(x^{3} + 64\right)\right) + 96\right)\right) - 60 = 0$$
or
$$\left(- 15 x + \left(\left(- 6 x^{2} + \left(x^{3} - \left(-4\right)^{3}\right)\right) + 6 \left(-4\right)^{2}\right)\right) - 60 = 0$$
$$- 15 \left(x + 4\right) + \left(- 6 \left(x^{2} - \left(-4\right)^{2}\right) + \left(x^{3} - \left(-4\right)^{3}\right)\right) = 0$$
$$- 15 \left(x + 4\right) + \left(\left(x - 4\right) \left(- 6 \left(x + 4\right)\right) + \left(x + 4\right) \left(\left(x^{2} - 4 x\right) + \left(-4\right)^{2}\right)\right) = 0$$
Take common factor 4 + x from the equation
we get:
$$\left(x + 4\right) \left(\left(- 6 \left(x - 4\right) + \left(\left(x^{2} - 4 x\right) + \left(-4\right)^{2}\right)\right) - 15\right) = 0$$
or
$$\left(x + 4\right) \left(x^{2} - 10 x + 25\right) = 0$$
then:
$$x_{1} = -4$$
and also
we get the equation
$$x^{2} - 10 x + 25 = 0$$
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
$$a = 1$$
$$b = -10$$
$$c = 25$$
, then
Because D = 0, then the equation has one root.
$$x_{2} = 5$$
The final answer for x^3 - 6*x^2 - 15*x + 100 = 0:
$$x_{1} = -4$$
$$x_{2} = 5$$
$$\left(- 15 x + \left(x^{3} - 6 x^{2}\right)\right) + 100 = 0$$
transform
$$\left(- 15 x + \left(\left(- 6 x^{2} + \left(x^{3} + 64\right)\right) + 96\right)\right) - 60 = 0$$
or
$$\left(- 15 x + \left(\left(- 6 x^{2} + \left(x^{3} - \left(-4\right)^{3}\right)\right) + 6 \left(-4\right)^{2}\right)\right) - 60 = 0$$
$$- 15 \left(x + 4\right) + \left(- 6 \left(x^{2} - \left(-4\right)^{2}\right) + \left(x^{3} - \left(-4\right)^{3}\right)\right) = 0$$
$$- 15 \left(x + 4\right) + \left(\left(x - 4\right) \left(- 6 \left(x + 4\right)\right) + \left(x + 4\right) \left(\left(x^{2} - 4 x\right) + \left(-4\right)^{2}\right)\right) = 0$$
Take common factor 4 + x from the equation
we get:
$$\left(x + 4\right) \left(\left(- 6 \left(x - 4\right) + \left(\left(x^{2} - 4 x\right) + \left(-4\right)^{2}\right)\right) - 15\right) = 0$$
or
$$\left(x + 4\right) \left(x^{2} - 10 x + 25\right) = 0$$
then:
$$x_{1} = -4$$
and also
we get the equation
$$x^{2} - 10 x + 25 = 0$$
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
$$a = 1$$
$$b = -10$$
$$c = 25$$
, then
D = b^2 - 4 * a * c =
(-10)^2 - 4 * (1) * (25) = 0
Because D = 0, then the equation has one root.
x = -b/2a = --10/2/(1)
$$x_{2} = 5$$
The final answer for x^3 - 6*x^2 - 15*x + 100 = 0:
$$x_{1} = -4$$
$$x_{2} = 5$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -6$$
$$q = \frac{c}{a}$$
$$q = -15$$
$$v = \frac{d}{a}$$
$$v = 100$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 6$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -15$$
$$x_{1} x_{2} x_{3} = 100$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -6$$
$$q = \frac{c}{a}$$
$$q = -15$$
$$v = \frac{d}{a}$$
$$v = 100$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 6$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -15$$
$$x_{1} x_{2} x_{3} = 100$$
Sum and product of roots
[src]
sum
-4 + 5
$$-4 + 5$$
=
1
$$1$$
product
-4*5
$$- 20$$
=
-20
$$-20$$
-20