Mister Exam
((1)/(sqrtx+1)-(sqrt2x-4))>0 inequation
The solution
You have entered
[src]
1 _____ --------- + - \/ 2*x + 4 > 0 ___ \/ x + 1
$$\left(4 - \sqrt{2 x}\right) + \frac{1}{\sqrt{x} + 1} > 0$$
4 - sqrt(2*x) + 1/(sqrt(x) + 1) > 0
Detail solution
Given the inequality:
$$\left(4 - \sqrt{2 x}\right) + \frac{1}{\sqrt{x} + 1} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(4 - \sqrt{2 x}\right) + \frac{1}{\sqrt{x} + 1} = 0$$
Solve:
$$x_{1} = \frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{9}{2}$$
$$x_{1} = \frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{9}{2}$$
This roots
$$x_{1} = \frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{9}{2}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{9}{2}\right)$$
=
$$\frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{22}{5}$$
substitute to the expression
$$\left(4 - \sqrt{2 x}\right) + \frac{1}{\sqrt{x} + 1} > 0$$
$$\left(4 - \sqrt{2 \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{22}{5}\right)}\right) + \frac{1}{1 + \sqrt{\frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{22}{5}}} > 0$$
the solution of our inequality is:
$$x < \frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{9}{2}$$
$$\left(4 - \sqrt{2 x}\right) + \frac{1}{\sqrt{x} + 1} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(4 - \sqrt{2 x}\right) + \frac{1}{\sqrt{x} + 1} = 0$$
Solve:
$$x_{1} = \frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{9}{2}$$
$$x_{1} = \frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{9}{2}$$
This roots
$$x_{1} = \frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{9}{2}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{9}{2}\right)$$
=
$$\frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{22}{5}$$
substitute to the expression
$$\left(4 - \sqrt{2 x}\right) + \frac{1}{\sqrt{x} + 1} > 0$$
$$\left(4 - \sqrt{2 \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{22}{5}\right)}\right) + \frac{1}{1 + \sqrt{\frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{22}{5}}} > 0$$
_________________________________
/ _______________
1 / 44 ___ / ___
4 + ------------------------------------------- - / -- + \/ 2 + \/ 33 + 18*\/ 2
_________________________________ \/ 5
/ _______________ > 0
/ ___ / ___
/ 22 \/ 2 \/ 33 + 18*\/ 2
1 + / -- + ----- + ------------------
\/ 5 2 2
the solution of our inequality is:
$$x < \frac{\sqrt{2}}{2} + \frac{\sqrt{18 \sqrt{2} + 33}}{2} + \frac{9}{2}$$
_____
\
-------ο-------
x1
Solving inequality on a graph
Rapid solution 2
[src]
______________
___ ___ / ___
9 \/ 2 \/ 3 *\/ 11 + 6*\/ 2
[0, - + ----- + -----------------------)
2 2 2
$$x\ in\ \left[0, \frac{\sqrt{2}}{2} + \frac{\sqrt{3} \sqrt{6 \sqrt{2} + 11}}{2} + \frac{9}{2}\right)$$
x in Interval.Ropen(0, sqrt(2)/2 + sqrt(3)*sqrt(6*sqrt(2) + 11)/2 + 9/2)
Rapid solution
[src]
/ ______________\ | ___ ___ / ___ | | 9 \/ 2 \/ 3 *\/ 11 + 6*\/ 2 | And|0 <= x, x < - + ----- + -----------------------| \ 2 2 2 /
$$0 \leq x \wedge x < \frac{\sqrt{2}}{2} + \frac{\sqrt{3} \sqrt{6 \sqrt{2} + 11}}{2} + \frac{9}{2}$$
(0 <= x)∧(x < 9/2 + sqrt(2)/2 + sqrt(3)*sqrt(11 + 6*sqrt(2))/2)