Mister Exam
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How to use it?
- Equation:
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Equation sin(x/3)=-1/2
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Equation x^3+4*x^2-x-4=0
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Equation 2^2-x-1=x^2-5*x-(-1-x^2)
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Equation 4(3y+1)^2-27=(4y+9)(4y-9)+2(5y+2)(2y-7)
- Express {x} in terms of y where:
- -14*x-16*y=14
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- Factor polynomial:
- x^3+4*x^2-x-4
-
Identical expressions
- x^ three + four *x^ two -x- four = zero
- x cubed plus 4 multiply by x squared minus x minus 4 equally 0
- x to the power of three plus four multiply by x to the power of two minus x minus four equally zero
- x3+4*x2-x-4=0
- x³+4*x²-x-4=0
- x to the power of 3+4*x to the power of 2-x-4=0
- x^3+4x^2-x-4=0
- x3+4x2-x-4=0
- x^3+4*x^2-x-4=O
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Similar expressions
- x^3+4*x^2-x+4=0
- x^3+4*x^2+x-4=0
- x^3-4*x^2-x-4=0
x^3+4*x^2-x-4=0 equation
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The solution
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3 2 x + 4*x - x - 4 = 0
$$\left(- x + \left(x^{3} + 4 x^{2}\right)\right) - 4 = 0$$
Detail solution
Given the equation:
$$\left(- x + \left(x^{3} + 4 x^{2}\right)\right) - 4 = 0$$
transform
$$\left(- x + \left(\left(4 x^{2} + \left(x^{3} - 1\right)\right) - 4\right)\right) + 1 = 0$$
or
$$\left(- x + \left(\left(4 x^{2} + \left(x^{3} - 1^{3}\right)\right) - 4 \cdot 1^{2}\right)\right) + 1 = 0$$
$$- (x - 1) + \left(4 \left(x^{2} - 1^{2}\right) + \left(x^{3} - 1^{3}\right)\right) = 0$$
$$- (x - 1) + \left(\left(x - 1\right) \left(\left(x^{2} + x\right) + 1^{2}\right) + 4 \left(x - 1\right) \left(x + 1\right)\right) = 0$$
Take common factor -1 + x from the equation
we get:
$$\left(x - 1\right) \left(\left(4 \left(x + 1\right) + \left(\left(x^{2} + x\right) + 1^{2}\right)\right) - 1\right) = 0$$
or
$$\left(x - 1\right) \left(x^{2} + 5 x + 4\right) = 0$$
then:
$$x_{1} = 1$$
and also
we get the equation
$$x^{2} + 5 x + 4 = 0$$
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
$$a = 1$$
$$b = 5$$
$$c = 4$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = -1$$
$$x_{3} = -4$$
The final answer for x^3 + 4*x^2 - x - 4 = 0:
$$x_{1} = 1$$
$$x_{2} = -1$$
$$x_{3} = -4$$
$$\left(- x + \left(x^{3} + 4 x^{2}\right)\right) - 4 = 0$$
transform
$$\left(- x + \left(\left(4 x^{2} + \left(x^{3} - 1\right)\right) - 4\right)\right) + 1 = 0$$
or
$$\left(- x + \left(\left(4 x^{2} + \left(x^{3} - 1^{3}\right)\right) - 4 \cdot 1^{2}\right)\right) + 1 = 0$$
$$- (x - 1) + \left(4 \left(x^{2} - 1^{2}\right) + \left(x^{3} - 1^{3}\right)\right) = 0$$
$$- (x - 1) + \left(\left(x - 1\right) \left(\left(x^{2} + x\right) + 1^{2}\right) + 4 \left(x - 1\right) \left(x + 1\right)\right) = 0$$
Take common factor -1 + x from the equation
we get:
$$\left(x - 1\right) \left(\left(4 \left(x + 1\right) + \left(\left(x^{2} + x\right) + 1^{2}\right)\right) - 1\right) = 0$$
or
$$\left(x - 1\right) \left(x^{2} + 5 x + 4\right) = 0$$
then:
$$x_{1} = 1$$
and also
we get the equation
$$x^{2} + 5 x + 4 = 0$$
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
$$a = 1$$
$$b = 5$$
$$c = 4$$
, then
D = b^2 - 4 * a * c =
(5)^2 - 4 * (1) * (4) = 9
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = -1$$
$$x_{3} = -4$$
The final answer for x^3 + 4*x^2 - x - 4 = 0:
$$x_{1} = 1$$
$$x_{2} = -1$$
$$x_{3} = -4$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 4$$
$$q = \frac{c}{a}$$
$$q = -1$$
$$v = \frac{d}{a}$$
$$v = -4$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = -4$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -1$$
$$x_{1} x_{2} x_{3} = -4$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 4$$
$$q = \frac{c}{a}$$
$$q = -1$$
$$v = \frac{d}{a}$$
$$v = -4$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = -4$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -1$$
$$x_{1} x_{2} x_{3} = -4$$
Sum and product of roots
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sum
-4 - 1 + 1
$$\left(-4 - 1\right) + 1$$
=
-4
$$-4$$
product
-4*(-1)
$$- -4$$
=
4
$$4$$
4
The graph