sqrt3+1

+ inequation

Solve the inequality!

A inequation with variable

The solution

You have entered [src]

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\/ 3  + 1 < x

$$1 + \sqrt{3} < x$$

1 + sqrt(3) < x

Detail solution

Given the inequality:
$$1 + \sqrt{3} < x$$
To solve this inequality, we must first solve the corresponding equation:
$$1 + \sqrt{3} = x$$
Solve:
Given the linear equation:

sqrt(3)+1 = x

Expand brackets in the left part

sqrt3+1 = x

Move free summands (without x)
from left part to right part, we given:
$$\sqrt{3} = x - 1$$
Move the summands with the unknown x
from the right part to the left part:
$$- x + \sqrt{3} = -1$$
Divide both parts of the equation by (sqrt(3) - x)/x

x = -1 / ((sqrt(3) - x)/x)

$$x_{1} = 1 + \sqrt{3}$$
$$x_{1} = 1 + \sqrt{3}$$
This roots
$$x_{1} = 1 + \sqrt{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \left(1 + \sqrt{3}\right)$$
=
$$\frac{9}{10} + \sqrt{3}$$
substitute to the expression
$$1 + \sqrt{3} < x$$
$$1 + \sqrt{3} < \frac{9}{10} + \sqrt{3}$$

      ___   9      ___
1 + \/ 3  < -- + \/ 3 
   10        

but

      ___   9      ___
1 + \/ 3  > -- + \/ 3 
   10        

Then
$$x < 1 + \sqrt{3}$$
no execute
the solution of our inequality is:
$$x > 1 + \sqrt{3}$$

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       x1

Solving inequality on a graph

Rapid solution [src]

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And\x < oo, 1 + \/ 3  < x/

$$x < \infty \wedge 1 + \sqrt{3} < x$$

(x < oo)∧(1 + sqrt(3) < x)

Rapid solution 2 [src]

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(1 + \/ 3 , oo)

$$x\ in\ \left(1 + \sqrt{3}, \infty\right)$$

x in Interval.open(1 + sqrt(3), oo)