Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 4-6*(x+2)=3-5*x
-
Equation 4-6(x+2)=3-5x
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Equation 1+sin(x)=0
-
Equation x^2-3*x+1=0
- Express {x} in terms of y where:
- 18*x-8*y=-2
- -11*x+15*y=7
- 6*x+17*y=3
- 1*x+10*y=-18
- Graphing y =:
- -4*x-x^3+4*x^2
-
Identical expressions
- - four *x-x^ three + four *x^ two = zero
- minus 4 multiply by x minus x cubed plus 4 multiply by x squared equally 0
- minus four multiply by x minus x to the power of three plus four multiply by x to the power of two equally zero
- -4*x-x3+4*x2=0
- -4*x-x³+4*x²=0
- -4*x-x to the power of 3+4*x to the power of 2=0
- -4x-x^3+4x^2=0
- -4x-x3+4x2=0
- -4*x-x^3+4*x^2=O
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Similar expressions
- -4*x-x^3-4*x^2=0
- -4*x+x^3+4*x^2=0
- 4*x-x^3+4*x^2=0
-4*x-x^3+4*x^2=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$4 x^{2} + \left(- x^{3} - 4 x\right) = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(- x^{2} + 4 x - 4\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$- x^{2} + 4 x - 4 = 0$$
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
$$a = -1$$
$$b = 4$$
$$c = -4$$
, then
Because D = 0, then the equation has one root.
$$x_{2} = 2$$
The final answer for -4*x - x^3 + 4*x^2 = 0:
$$x_{1} = 0$$
$$x_{2} = 2$$
$$4 x^{2} + \left(- x^{3} - 4 x\right) = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(- x^{2} + 4 x - 4\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$- x^{2} + 4 x - 4 = 0$$
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
$$a = -1$$
$$b = 4$$
$$c = -4$$
, then
D = b^2 - 4 * a * c =
(4)^2 - 4 * (-1) * (-4) = 0
Because D = 0, then the equation has one root.
x = -b/2a = -4/2/(-1)
$$x_{2} = 2$$
The final answer for -4*x - x^3 + 4*x^2 = 0:
$$x_{1} = 0$$
$$x_{2} = 2$$
Vieta's Theorem
rewrite the equation
$$4 x^{2} + \left(- x^{3} - 4 x\right) = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - 4 x^{2} + 4 x = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -4$$
$$q = \frac{c}{a}$$
$$q = 4$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 4$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 4$$
$$x_{1} x_{2} x_{3} = 0$$
$$4 x^{2} + \left(- x^{3} - 4 x\right) = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - 4 x^{2} + 4 x = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -4$$
$$q = \frac{c}{a}$$
$$q = 4$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 4$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 4$$
$$x_{1} x_{2} x_{3} = 0$$
The graph