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Similar expressions
- x^3+11x^2+9x-99=0
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x^3+11x^2-9x-99=0 equation
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The solution
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3 2 x + 11*x - 9*x - 99 = 0
$$\left(- 9 x + \left(x^{3} + 11 x^{2}\right)\right) - 99 = 0$$
Detail solution
Given the equation:
$$\left(- 9 x + \left(x^{3} + 11 x^{2}\right)\right) - 99 = 0$$
transform
$$\left(- 9 x + \left(\left(11 x^{2} + \left(x^{3} - 27\right)\right) - 99\right)\right) + 27 = 0$$
or
$$\left(- 9 x + \left(\left(11 x^{2} + \left(x^{3} - 3^{3}\right)\right) - 11 \cdot 3^{2}\right)\right) + 3 \cdot 9 = 0$$
$$- 9 \left(x - 3\right) + \left(11 \left(x^{2} - 3^{2}\right) + \left(x^{3} - 3^{3}\right)\right) = 0$$
$$- 9 \left(x - 3\right) + \left(\left(x - 3\right) \left(\left(x^{2} + 3 x\right) + 3^{2}\right) + 11 \left(x - 3\right) \left(x + 3\right)\right) = 0$$
Take common factor -3 + x from the equation
we get:
$$\left(x - 3\right) \left(\left(11 \left(x + 3\right) + \left(\left(x^{2} + 3 x\right) + 3^{2}\right)\right) - 9\right) = 0$$
or
$$\left(x - 3\right) \left(x^{2} + 14 x + 33\right) = 0$$
then:
$$x_{1} = 3$$
and also
we get the equation
$$x^{2} + 14 x + 33 = 0$$
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
$$a = 1$$
$$b = 14$$
$$c = 33$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = -3$$
$$x_{3} = -11$$
The final answer for x^3 + 11*x^2 - 9*x - 99 = 0:
$$x_{1} = 3$$
$$x_{2} = -3$$
$$x_{3} = -11$$
$$\left(- 9 x + \left(x^{3} + 11 x^{2}\right)\right) - 99 = 0$$
transform
$$\left(- 9 x + \left(\left(11 x^{2} + \left(x^{3} - 27\right)\right) - 99\right)\right) + 27 = 0$$
or
$$\left(- 9 x + \left(\left(11 x^{2} + \left(x^{3} - 3^{3}\right)\right) - 11 \cdot 3^{2}\right)\right) + 3 \cdot 9 = 0$$
$$- 9 \left(x - 3\right) + \left(11 \left(x^{2} - 3^{2}\right) + \left(x^{3} - 3^{3}\right)\right) = 0$$
$$- 9 \left(x - 3\right) + \left(\left(x - 3\right) \left(\left(x^{2} + 3 x\right) + 3^{2}\right) + 11 \left(x - 3\right) \left(x + 3\right)\right) = 0$$
Take common factor -3 + x from the equation
we get:
$$\left(x - 3\right) \left(\left(11 \left(x + 3\right) + \left(\left(x^{2} + 3 x\right) + 3^{2}\right)\right) - 9\right) = 0$$
or
$$\left(x - 3\right) \left(x^{2} + 14 x + 33\right) = 0$$
then:
$$x_{1} = 3$$
and also
we get the equation
$$x^{2} + 14 x + 33 = 0$$
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
$$a = 1$$
$$b = 14$$
$$c = 33$$
, then
D = b^2 - 4 * a * c =
(14)^2 - 4 * (1) * (33) = 64
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = -3$$
$$x_{3} = -11$$
The final answer for x^3 + 11*x^2 - 9*x - 99 = 0:
$$x_{1} = 3$$
$$x_{2} = -3$$
$$x_{3} = -11$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 11$$
$$q = \frac{c}{a}$$
$$q = -9$$
$$v = \frac{d}{a}$$
$$v = -99$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = -11$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -9$$
$$x_{1} x_{2} x_{3} = -99$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 11$$
$$q = \frac{c}{a}$$
$$q = -9$$
$$v = \frac{d}{a}$$
$$v = -99$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = -11$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -9$$
$$x_{1} x_{2} x_{3} = -99$$
Sum and product of roots
[src]
sum
-11 - 3 + 3
$$\left(-11 - 3\right) + 3$$
=
-11
$$-11$$
product
-11*(-3)*3
$$3 \left(- -33\right)$$
=
99
$$99$$
99