Mister Exam
(4-x)√4+3x-x²≤0 inequation
The solution
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___ 2 (4 - x)*\/ 4 + 3*x - x <= 0
$$- x^{2} + \left(3 x + \sqrt{4} \left(4 - x\right)\right) \leq 0$$
-x^2 + 3*x + sqrt(4)*(4 - x) <= 0
Detail solution
Given the inequality:
$$- x^{2} + \left(3 x + \sqrt{4} \left(4 - x\right)\right) \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$- x^{2} + \left(3 x + \sqrt{4} \left(4 - x\right)\right) = 0$$
Solve:
Expand the expression in the equation
$$- x^{2} + \left(3 x + \sqrt{4} \left(4 - x\right)\right) = 0$$
We get the quadratic equation
$$- x^{2} + x + 8 = 0$$
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
$$a = -1$$
$$b = 1$$
$$c = 8$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = \frac{1}{2} - \frac{\sqrt{33}}{2}$$
$$x_{2} = \frac{1}{2} + \frac{\sqrt{33}}{2}$$
$$x_{1} = \frac{1}{2} - \frac{\sqrt{33}}{2}$$
$$x_{2} = \frac{1}{2} + \frac{\sqrt{33}}{2}$$
$$x_{1} = \frac{1}{2} - \frac{\sqrt{33}}{2}$$
$$x_{2} = \frac{1}{2} + \frac{\sqrt{33}}{2}$$
This roots
$$x_{1} = \frac{1}{2} - \frac{\sqrt{33}}{2}$$
$$x_{2} = \frac{1}{2} + \frac{\sqrt{33}}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{1}{2} - \frac{\sqrt{33}}{2}\right) + - \frac{1}{10}$$
=
$$\frac{2}{5} - \frac{\sqrt{33}}{2}$$
substitute to the expression
$$- x^{2} + \left(3 x + \sqrt{4} \left(4 - x\right)\right) \leq 0$$
$$- \left(\frac{2}{5} - \frac{\sqrt{33}}{2}\right)^{2} + \left(3 \left(\frac{2}{5} - \frac{\sqrt{33}}{2}\right) + \sqrt{4} \left(4 - \left(\frac{2}{5} - \frac{\sqrt{33}}{2}\right)\right)\right) \leq 0$$
one of the solutions of our inequality is:
$$x \leq \frac{1}{2} - \frac{\sqrt{33}}{2}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq \frac{1}{2} - \frac{\sqrt{33}}{2}$$
$$x \geq \frac{1}{2} + \frac{\sqrt{33}}{2}$$
$$- x^{2} + \left(3 x + \sqrt{4} \left(4 - x\right)\right) \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$- x^{2} + \left(3 x + \sqrt{4} \left(4 - x\right)\right) = 0$$
Solve:
Expand the expression in the equation
$$- x^{2} + \left(3 x + \sqrt{4} \left(4 - x\right)\right) = 0$$
We get the quadratic equation
$$- x^{2} + x + 8 = 0$$
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
$$a = -1$$
$$b = 1$$
$$c = 8$$
, then
D = b^2 - 4 * a * c =
(1)^2 - 4 * (-1) * (8) = 33
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{1}{2} - \frac{\sqrt{33}}{2}$$
$$x_{2} = \frac{1}{2} + \frac{\sqrt{33}}{2}$$
$$x_{1} = \frac{1}{2} - \frac{\sqrt{33}}{2}$$
$$x_{2} = \frac{1}{2} + \frac{\sqrt{33}}{2}$$
$$x_{1} = \frac{1}{2} - \frac{\sqrt{33}}{2}$$
$$x_{2} = \frac{1}{2} + \frac{\sqrt{33}}{2}$$
This roots
$$x_{1} = \frac{1}{2} - \frac{\sqrt{33}}{2}$$
$$x_{2} = \frac{1}{2} + \frac{\sqrt{33}}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{1}{2} - \frac{\sqrt{33}}{2}\right) + - \frac{1}{10}$$
=
$$\frac{2}{5} - \frac{\sqrt{33}}{2}$$
substitute to the expression
$$- x^{2} + \left(3 x + \sqrt{4} \left(4 - x\right)\right) \leq 0$$
$$- \left(\frac{2}{5} - \frac{\sqrt{33}}{2}\right)^{2} + \left(3 \left(\frac{2}{5} - \frac{\sqrt{33}}{2}\right) + \sqrt{4} \left(4 - \left(\frac{2}{5} - \frac{\sqrt{33}}{2}\right)\right)\right) \leq 0$$
2
/ ____\ ____
42 |2 \/ 33 | \/ 33 <= 0
-- - |- - ------| - ------
5 \5 2 / 2 one of the solutions of our inequality is:
$$x \leq \frac{1}{2} - \frac{\sqrt{33}}{2}$$
_____ _____
\ /
-------•-------•-------
x1 x2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq \frac{1}{2} - \frac{\sqrt{33}}{2}$$
$$x \geq \frac{1}{2} + \frac{\sqrt{33}}{2}$$
Solving inequality on a graph
Rapid solution
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/ / ____ \ / ____ \\ | | 1 \/ 33 | |1 \/ 33 || Or|And|x <= - - ------, -oo < x|, And|- + ------ <= x, x < oo|| \ \ 2 2 / \2 2 //
$$\left(x \leq \frac{1}{2} - \frac{\sqrt{33}}{2} \wedge -\infty < x\right) \vee \left(\frac{1}{2} + \frac{\sqrt{33}}{2} \leq x \wedge x < \infty\right)$$
((-oo < x)∧(x <= 1/2 - sqrt(33)/2))∨((x < oo)∧(1/2 + sqrt(33)/2 <= x))
Rapid solution 2
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____ ____
1 \/ 33 1 \/ 33
(-oo, - - ------] U [- + ------, oo)
2 2 2 2
$$x\ in\ \left(-\infty, \frac{1}{2} - \frac{\sqrt{33}}{2}\right] \cup \left[\frac{1}{2} + \frac{\sqrt{33}}{2}, \infty\right)$$
x in Union(Interval(-oo, 1/2 - sqrt(33)/2), Interval(1/2 + sqrt(33)/2, oo))