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- Square Inequality Step-by-Step
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- Logarithmic inequality online
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How to use it?
- Inequation:
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((2-x^2)*(x-3)^3)/((x+1)*(x^2-3x-4))>=0
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(31-5*2^x)*1/(4^x-24*2^x+128)>=0,25
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6x^2+17<0
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ctgx>=sqrt(3)/3
- Canonical form:
- =0
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Identical expressions
- ((two -x^ two)*(x- three)^ three)/((x+ one)*(x^ two -3x- four))>= zero
- ((2 minus x squared ) multiply by (x minus 3) cubed ) divide by ((x plus 1) multiply by (x squared minus 3x minus 4)) greater than or equal to 0
- ((two minus x to the power of two) multiply by (x minus three) to the power of three) divide by ((x plus one) multiply by (x to the power of two minus 3x minus four)) greater than or equal to zero
- ((2-x2)*(x-3)3)/((x+1)*(x2-3x-4))>=0
- 2-x2*x-33/x+1*x2-3x-4>=0
- ((2-x²)*(x-3)³)/((x+1)*(x²-3x-4))>=0
- ((2-x to the power of 2)*(x-3) to the power of 3)/((x+1)*(x to the power of 2-3x-4))>=0
- ((2-x^2)(x-3)^3)/((x+1)(x^2-3x-4))>=0
- ((2-x2)(x-3)3)/((x+1)(x2-3x-4))>=0
- 2-x2x-33/x+1x2-3x-4>=0
- 2-x^2x-3^3/x+1x^2-3x-4>=0
- ((2-x^2)*(x-3)^3)/((x+1)*(x^2-3x-4))>=O
- ((2-x^2)*(x-3)^3) divide by ((x+1)*(x^2-3x-4))>=0
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Similar expressions
- ((2-x^2)*(x-3)^3)/((x+1)*(x^2+3x-4))>=0
- ((2-x^2)*(x-3)^3)/((x-1)*(x^2-3x-4))>=0
- ((2-x^2)*(x-3)^3)/((x+1)*(x^2-3x+4))>=0
- ((2-x^2)*(x+3)^3)/((x+1)*(x^2-3x-4))>=0
- ((2+x^2)*(x-3)^3)/((x+1)*(x^2-3x-4))>=0
((2-x^2)*(x-3)^3)/((x+1)*(x^2-3x-4))>=0 inequation
The solution
You have entered
[src]
/ 2\ 3
\2 - x /*(x - 3)
---------------------- >= 0
/ 2 \
(x + 1)*\x - 3*x - 4/
$$\frac{\left(2 - x^{2}\right) \left(x - 3\right)^{3}}{\left(x + 1\right) \left(x^{2} - 3 x - 4\right)} \geq 0$$
(2 - x^2)*(x - 1*3)^3/(((x + 1)*(x^2 - 3*x - 1*4))) >= 0
Detail solution
Given the inequality:
$$\frac{\left(2 - x^{2}\right) \left(x - 3\right)^{3}}{\left(x + 1\right) \left(x^{2} - 3 x - 4\right)} \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(2 - x^{2}\right) \left(x - 3\right)^{3}}{\left(x + 1\right) \left(x^{2} - 3 x - 4\right)} = 0$$
Solve:
$$x_{1} = 3$$
$$x_{2} = - \sqrt{2}$$
$$x_{3} = \sqrt{2}$$
$$x_{1} = 3$$
$$x_{2} = - \sqrt{2}$$
$$x_{3} = \sqrt{2}$$
This roots
$$x_{2} = - \sqrt{2}$$
$$x_{3} = \sqrt{2}$$
$$x_{1} = 3$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \sqrt{2} - \frac{1}{10}$$
=
$$- \sqrt{2} - \frac{1}{10}$$
substitute to the expression
$$\frac{\left(2 - x^{2}\right) \left(x - 3\right)^{3}}{\left(x + 1\right) \left(x^{2} - 3 x - 4\right)} \geq 0$$
$$\frac{\left(2 - \left(- \sqrt{2} - \frac{1}{10}\right)^{2}\right) \left(\left(-1\right) 3 - \left(\frac{1}{10} + \sqrt{2}\right)\right)^{3}}{\left(\left(- \sqrt{2} - \frac{1}{10}\right) + 1\right) \left(\left(-1\right) 4 + \left(- \sqrt{2} - \frac{1}{10}\right)^{2} - 3 \left(- \sqrt{2} - \frac{1}{10}\right)\right)} \geq 0$$
but
Then
$$x \leq - \sqrt{2}$$
no execute
one of the solutions of our inequality is:
$$x \geq - \sqrt{2} \wedge x \leq \sqrt{2}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \geq - \sqrt{2} \wedge x \leq \sqrt{2}$$
$$x \geq 3$$
$$\frac{\left(2 - x^{2}\right) \left(x - 3\right)^{3}}{\left(x + 1\right) \left(x^{2} - 3 x - 4\right)} \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(2 - x^{2}\right) \left(x - 3\right)^{3}}{\left(x + 1\right) \left(x^{2} - 3 x - 4\right)} = 0$$
Solve:
$$x_{1} = 3$$
$$x_{2} = - \sqrt{2}$$
$$x_{3} = \sqrt{2}$$
$$x_{1} = 3$$
$$x_{2} = - \sqrt{2}$$
$$x_{3} = \sqrt{2}$$
This roots
$$x_{2} = - \sqrt{2}$$
$$x_{3} = \sqrt{2}$$
$$x_{1} = 3$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \sqrt{2} - \frac{1}{10}$$
=
$$- \sqrt{2} - \frac{1}{10}$$
substitute to the expression
$$\frac{\left(2 - x^{2}\right) \left(x - 3\right)^{3}}{\left(x + 1\right) \left(x^{2} - 3 x - 4\right)} \geq 0$$
$$\frac{\left(2 - \left(- \sqrt{2} - \frac{1}{10}\right)^{2}\right) \left(\left(-1\right) 3 - \left(\frac{1}{10} + \sqrt{2}\right)\right)^{3}}{\left(\left(- \sqrt{2} - \frac{1}{10}\right) + 1\right) \left(\left(-1\right) 4 + \left(- \sqrt{2} - \frac{1}{10}\right)^{2} - 3 \left(- \sqrt{2} - \frac{1}{10}\right)\right)} \geq 0$$
3 / 2\
/ 31 ___\ | / 1 ___\ |
|- -- - \/ 2 | *|2 - |- -- - \/ 2 | |
\ 10 / \ \ 10 / /
----------------------------------------------- >= 0
/ 2 \
/9 ___\ | 37 / 1 ___\ ___|
|-- - \/ 2 |*|- -- + |- -- - \/ 2 | + 3*\/ 2 |
\10 / \ 10 \ 10 / / but
3 / 2\
/ 31 ___\ | / 1 ___\ |
|- -- - \/ 2 | *|2 - |- -- - \/ 2 | |
\ 10 / \ \ 10 / /
----------------------------------------------- < 0
/ 2 \
/9 ___\ | 37 / 1 ___\ ___|
|-- - \/ 2 |*|- -- + |- -- - \/ 2 | + 3*\/ 2 |
\10 / \ 10 \ 10 / / Then
$$x \leq - \sqrt{2}$$
no execute
one of the solutions of our inequality is:
$$x \geq - \sqrt{2} \wedge x \leq \sqrt{2}$$
_____ _____
/ \ /
-------•-------•-------•-------
x_2 x_3 x_1Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \geq - \sqrt{2} \wedge x \leq \sqrt{2}$$
$$x \geq 3$$
Solving inequality on a graph
Rapid solution
[src]
/ / ___ \ / ___ \\ Or\And(3 <= x, x < 4), And\x <= \/ 2 , -1 < x/, And\-\/ 2 <= x, x < -1//
$$\left(3 \leq x \wedge x < 4\right) \vee \left(x \leq \sqrt{2} \wedge -1 < x\right) \vee \left(- \sqrt{2} \leq x \wedge x < -1\right)$$
((3 <= x)∧(x < 4))∨((-1 < x)∧(x <= sqrt(2)))∨((x < -1)∧(-sqrt(2) <= x))
Rapid solution 2
[src]
___ ___ [-\/ 2 , -1) U (-1, \/ 2 ] U [3, 4)
$$x\ in\ \left[- \sqrt{2}, -1\right) \cup \left(-1, \sqrt{2}\right] \cup \left[3, 4\right)$$
x in Union(Interval.Lopen(-1, sqrt(2)), Interval.Ropen(3, 4), Interval.Ropen(-sqrt(2), -1))
The graph