Mister Exam
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How to use it?
- Equation:
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Equation 7*x=0
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Equation 3x-2=x+4
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Equation tan(x)=7
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Equation (x-3)(x^2-2x+1)=3(x-1)
- Express {x} in terms of y where:
- 18*x-10*y=-9
- 18*x-12*y=-9
- 12*x+11*y=20
- 19*x-12*y=-9
- Graphing y =:
- 2*x^3+3*x^2-1
- Derivative of:
- 2*x^3+3*x^2-1
-
Identical expressions
- two *x^ three + three *x^ two - one = zero
- 2 multiply by x cubed plus 3 multiply by x squared minus 1 equally 0
- two multiply by x to the power of three plus three multiply by x to the power of two minus one equally zero
- 2*x3+3*x2-1=0
- 2*x³+3*x²-1=0
- 2*x to the power of 3+3*x to the power of 2-1=0
- 2x^3+3x^2-1=0
- 2x3+3x2-1=0
- 2*x^3+3*x^2-1=O
-
Similar expressions
- 2*x^3-3*x^2-1=0
- 2*x^3+3*x^2+1=0
2*x^3+3*x^2-1=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\left(2 x^{3} + 3 x^{2}\right) - 1 = 0$$
transform
$$\left(3 x^{2} + \left(2 x^{3} + 2\right)\right) - 3 = 0$$
or
$$\left(3 x^{2} + \left(2 x^{3} - 2 \left(-1\right)^{3}\right)\right) - 3 \left(-1\right)^{2} = 0$$
$$3 \left(x^{2} - \left(-1\right)^{2}\right) + 2 \left(x^{3} - \left(-1\right)^{3}\right) = 0$$
$$\left(x - 1\right) 3 \left(x + 1\right) + 2 \left(x + 1\right) \left(\left(x^{2} - x\right) + \left(-1\right)^{2}\right) = 0$$
Take common factor 1 + x from the equation
we get:
$$\left(x + 1\right) \left(3 \left(x - 1\right) + 2 \left(\left(x^{2} - x\right) + \left(-1\right)^{2}\right)\right) = 0$$
or
$$\left(x + 1\right) \left(2 x^{2} + x - 1\right) = 0$$
then:
$$x_{1} = -1$$
and also
we get the equation
$$2 x^{2} + x - 1 = 0$$
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
$$a = 2$$
$$b = 1$$
$$c = -1$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = \frac{1}{2}$$
$$x_{3} = -1$$
The final answer for 2*x^3 + 3*x^2 - 1 = 0:
$$x_{1} = -1$$
$$x_{2} = \frac{1}{2}$$
$$x_{3} = -1$$
$$\left(2 x^{3} + 3 x^{2}\right) - 1 = 0$$
transform
$$\left(3 x^{2} + \left(2 x^{3} + 2\right)\right) - 3 = 0$$
or
$$\left(3 x^{2} + \left(2 x^{3} - 2 \left(-1\right)^{3}\right)\right) - 3 \left(-1\right)^{2} = 0$$
$$3 \left(x^{2} - \left(-1\right)^{2}\right) + 2 \left(x^{3} - \left(-1\right)^{3}\right) = 0$$
$$\left(x - 1\right) 3 \left(x + 1\right) + 2 \left(x + 1\right) \left(\left(x^{2} - x\right) + \left(-1\right)^{2}\right) = 0$$
Take common factor 1 + x from the equation
we get:
$$\left(x + 1\right) \left(3 \left(x - 1\right) + 2 \left(\left(x^{2} - x\right) + \left(-1\right)^{2}\right)\right) = 0$$
or
$$\left(x + 1\right) \left(2 x^{2} + x - 1\right) = 0$$
then:
$$x_{1} = -1$$
and also
we get the equation
$$2 x^{2} + x - 1 = 0$$
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
$$a = 2$$
$$b = 1$$
$$c = -1$$
, then
D = b^2 - 4 * a * c =
(1)^2 - 4 * (2) * (-1) = 9
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = \frac{1}{2}$$
$$x_{3} = -1$$
The final answer for 2*x^3 + 3*x^2 - 1 = 0:
$$x_{1} = -1$$
$$x_{2} = \frac{1}{2}$$
$$x_{3} = -1$$
Vieta's Theorem
rewrite the equation
$$\left(2 x^{3} + 3 x^{2}\right) - 1 = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} + \frac{3 x^{2}}{2} - \frac{1}{2} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{3}{2}$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = - \frac{1}{2}$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = - \frac{3}{2}$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = - \frac{1}{2}$$
$$\left(2 x^{3} + 3 x^{2}\right) - 1 = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} + \frac{3 x^{2}}{2} - \frac{1}{2} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{3}{2}$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = - \frac{1}{2}$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = - \frac{3}{2}$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = - \frac{1}{2}$$
Sum and product of roots
[src]
sum
-1 + 1/2
$$-1 + \frac{1}{2}$$
=
-1/2
$$- \frac{1}{2}$$
product
-1 --- 2
$$- \frac{1}{2}$$
=
-1/2
$$- \frac{1}{2}$$
-1/2
The graph