Mister Exam
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How to use it?
- Equation:
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Equation t-4/25=8
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Equation x^2+5*x-14=0
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Equation x^2=5
- Express {x} in terms of y where:
- 18*x+20*y=-13
- -19*x-16*y=3
- 19*x-5*y=6
- 11*x-10*y=-19
- Limit of the function:
-
x^3-36*x
-
Identical expressions
- x^ three - thirty-six *x= zero
- x cubed minus 36 multiply by x equally 0
- x to the power of three minus thirty minus six multiply by x equally zero
- x3-36*x=0
- x³-36*x=0
- x to the power of 3-36*x=0
- x^3-36x=0
- x3-36x=0
- x^3-36*x=O
-
Similar expressions
- x^3+36*x=0
x^3-36*x=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x^{3} - 36 x = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(x^{2} - 36\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$x^{2} - 36 = 0$$
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
$$a = 1$$
$$b = 0$$
$$c = -36$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = 6$$
$$x_{3} = -6$$
The final answer for x^3 - 36*x = 0:
$$x_{1} = 0$$
$$x_{2} = 6$$
$$x_{3} = -6$$
$$x^{3} - 36 x = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(x^{2} - 36\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$x^{2} - 36 = 0$$
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
$$a = 1$$
$$b = 0$$
$$c = -36$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (1) * (-36) = 144
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = 6$$
$$x_{3} = -6$$
The final answer for x^3 - 36*x = 0:
$$x_{1} = 0$$
$$x_{2} = 6$$
$$x_{3} = -6$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -36$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -36$$
$$x_{1} x_{2} x_{3} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -36$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -36$$
$$x_{1} x_{2} x_{3} = 0$$
Sum and product of roots
[src]
sum
-6 + 6
$$-6 + 6$$
=
0
$$0$$
product
-6*0*6
$$6 \left(- 0\right)$$
=
0
$$0$$
0