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How to use it?
- Inequation:
- (5-4x)/(3x^2-x-4)<4
- 4(x+5)-40≥-5+x
-
tgx>=-1
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8x²+24x>=0
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Identical expressions
- (five - four x)/(3x^ two -x- four)<4
- (5 minus 4x) divide by (3x squared minus x minus 4) less than 4
- (five minus four x) divide by (3x to the power of two minus x minus four) less than 4
- (5-4x)/(3x2-x-4)<4
- 5-4x/3x2-x-4<4
- (5-4x)/(3x²-x-4)<4
- (5-4x)/(3x to the power of 2-x-4)<4
- 5-4x/3x^2-x-4<4
- (5-4x) divide by (3x^2-x-4)<4
-
Similar expressions
- (5-4x)/(3x^2-x+4)<4
- (5+4x)/(3x^2-x-4)<4
- (5-4x)/(3x^2+x-4)<4
(5-4x)/(3x^2-x-4)<4 inequation
The solution
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[src]
5 - 4*x ------------ < 4 2 3*x - x - 4
$$\frac{5 - 4 x}{\left(3 x^{2} - x\right) - 4} < 4$$
(5 - 4*x)/(3*x^2 - x - 4) < 4
Detail solution
Given the inequality:
$$\frac{5 - 4 x}{\left(3 x^{2} - x\right) - 4} < 4$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{5 - 4 x}{\left(3 x^{2} - x\right) - 4} = 4$$
Solve:
Given the equation:
$$\frac{5 - 4 x}{\left(3 x^{2} - x\right) - 4} = 4$$
Multiply the equation sides by the denominators:
-4 - x + 3*x^2
we get:
$$\frac{\left(5 - 4 x\right) \left(3 x^{2} - x - 4\right)}{\left(3 x^{2} - x\right) - 4} = 12 x^{2} - 4 x - 16$$
$$5 - 4 x = 12 x^{2} - 4 x - 16$$
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$5 - 4 x = 12 x^{2} - 4 x - 16$$
to
$$21 - 12 x^{2} = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -12$$
$$b = 0$$
$$c = 21$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = - \frac{\sqrt{7}}{2}$$
$$x_{2} = \frac{\sqrt{7}}{2}$$
$$x_{1} = - \frac{\sqrt{7}}{2}$$
$$x_{2} = \frac{\sqrt{7}}{2}$$
$$x_{1} = - \frac{\sqrt{7}}{2}$$
$$x_{2} = \frac{\sqrt{7}}{2}$$
This roots
$$x_{1} = - \frac{\sqrt{7}}{2}$$
$$x_{2} = \frac{\sqrt{7}}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\sqrt{7}}{2} - \frac{1}{10}$$
=
$$- \frac{\sqrt{7}}{2} - \frac{1}{10}$$
substitute to the expression
$$\frac{5 - 4 x}{\left(3 x^{2} - x\right) - 4} < 4$$
$$\frac{5 - 4 \left(- \frac{\sqrt{7}}{2} - \frac{1}{10}\right)}{-4 + \left(- (- \frac{\sqrt{7}}{2} - \frac{1}{10}) + 3 \left(- \frac{\sqrt{7}}{2} - \frac{1}{10}\right)^{2}\right)} < 4$$
one of the solutions of our inequality is:
$$x < - \frac{\sqrt{7}}{2}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \frac{\sqrt{7}}{2}$$
$$x > \frac{\sqrt{7}}{2}$$
$$\frac{5 - 4 x}{\left(3 x^{2} - x\right) - 4} < 4$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{5 - 4 x}{\left(3 x^{2} - x\right) - 4} = 4$$
Solve:
Given the equation:
$$\frac{5 - 4 x}{\left(3 x^{2} - x\right) - 4} = 4$$
Multiply the equation sides by the denominators:
-4 - x + 3*x^2
we get:
$$\frac{\left(5 - 4 x\right) \left(3 x^{2} - x - 4\right)}{\left(3 x^{2} - x\right) - 4} = 12 x^{2} - 4 x - 16$$
$$5 - 4 x = 12 x^{2} - 4 x - 16$$
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$5 - 4 x = 12 x^{2} - 4 x - 16$$
to
$$21 - 12 x^{2} = 0$$
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:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -12$$
$$b = 0$$
$$c = 21$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (-12) * (21) = 1008
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = - \frac{\sqrt{7}}{2}$$
$$x_{2} = \frac{\sqrt{7}}{2}$$
$$x_{1} = - \frac{\sqrt{7}}{2}$$
$$x_{2} = \frac{\sqrt{7}}{2}$$
$$x_{1} = - \frac{\sqrt{7}}{2}$$
$$x_{2} = \frac{\sqrt{7}}{2}$$
This roots
$$x_{1} = - \frac{\sqrt{7}}{2}$$
$$x_{2} = \frac{\sqrt{7}}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\sqrt{7}}{2} - \frac{1}{10}$$
=
$$- \frac{\sqrt{7}}{2} - \frac{1}{10}$$
substitute to the expression
$$\frac{5 - 4 x}{\left(3 x^{2} - x\right) - 4} < 4$$
$$\frac{5 - 4 \left(- \frac{\sqrt{7}}{2} - \frac{1}{10}\right)}{-4 + \left(- (- \frac{\sqrt{7}}{2} - \frac{1}{10}) + 3 \left(- \frac{\sqrt{7}}{2} - \frac{1}{10}\right)^{2}\right)} < 4$$
27 ___
-- + 2*\/ 7
5
--------------------------------
2 < 4
___ / ___\
39 \/ 7 | 1 \/ 7 |
- -- + ----- + 3*|- -- - -----|
10 2 \ 10 2 / one of the solutions of our inequality is:
$$x < - \frac{\sqrt{7}}{2}$$
_____ _____
\ /
-------ο-------ο-------
x1 x2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \frac{\sqrt{7}}{2}$$
$$x > \frac{\sqrt{7}}{2}$$
Solving inequality on a graph
Rapid solution
[src]
/ / ___ \ / ___\ \ | | -\/ 7 | | \/ 7 | | Or|And|-oo < x, x < -------|, And|-1 < x, x < -----|, And(4/3 < x, x < oo)| \ \ 2 / \ 2 / /
$$\left(-\infty < x \wedge x < - \frac{\sqrt{7}}{2}\right) \vee \left(-1 < x \wedge x < \frac{\sqrt{7}}{2}\right) \vee \left(\frac{4}{3} < x \wedge x < \infty\right)$$
((4/3 < x)∧(x < oo))∨((-oo < x)∧(x < -sqrt(7)/2))∨((-1 < x)∧(x < sqrt(7)/2))
Rapid solution 2
[src]
___ ___
-\/ 7 \/ 7
(-oo, -------) U (-1, -----) U (4/3, oo)
2 2
$$x\ in\ \left(-\infty, - \frac{\sqrt{7}}{2}\right) \cup \left(-1, \frac{\sqrt{7}}{2}\right) \cup \left(\frac{4}{3}, \infty\right)$$
x in Union(Interval.open(-oo, -sqrt(7)/2), Interval.open(-1, sqrt(7)/2), Interval.open(4/3, oo))