sqrt(x+78)

+ inequation

Solve the inequality!

A inequation with variable

The solution

You have entered [src]

  ________        
\/ x + 78  < x + 6

$$\sqrt{x + 78} < x + 6$$

sqrt(x + 78) < x + 6

Detail solution

Given the inequality:
$$\sqrt{x + 78} < x + 6$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{x + 78} = x + 6$$
Solve:
Given the equation
$$\sqrt{x + 78} = x + 6$$
$$\sqrt{x + 78} = x + 6$$
We raise the equation sides to 2-th degree
$$x + 78 = \left(x + 6\right)^{2}$$
$$x + 78 = x^{2} + 12 x + 36$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} - 11 x + 42 = 0$$
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
$$a = -1$$
$$b = -11$$
$$c = 42$$
, then

D = b^2 - 4 * a * c = 

(-11)^2 - 4 * (-1) * (42) = 289

Because D > 0, then the equation has two roots.

x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
$$x_{1} = -14$$
Simplify
$$x_{2} = 3$$
Simplify

Because
$$\sqrt{x + 78} = x + 6$$
and
$$\sqrt{x + 78} \geq 0$$
then
$$x + 6 \geq 0$$
or
$$-6 \leq x$$
$$x < \infty$$
$$x_{2} = 3$$
$$x_{1} = 3$$
$$x_{1} = 3$$
This roots
$$x_{1} = 3$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 3$$
=
$$\frac{29}{10}$$
substitute to the expression
$$\sqrt{x + 78} < x + 6$$
$$\sqrt{\frac{29}{10} + 78} < \frac{29}{10} + 6$$

  ______     
\/ 8090    89
-------- < --
   10      10
     

but

  ______     
\/ 8090    89
-------- > --
   10      10
     

Then
$$x < 3$$
no execute
the solution of our inequality is:
$$x > 3$$

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        /
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       x_1

Solving inequality on a graph

Rapid solution [src]

And(3 < x, x < oo)

$$3 < x \wedge x < \infty$$

(3 < x)∧(x < oo)

Rapid solution 2 [src]

(3, oo)

$$x\ in\ \left(3, \infty\right)$$

x in Interval.open(3, oo)

The graph