Solve the inequality x^2+2x+6>=0.
x^2 + 2x + 6 >= 0
To solve, we will need to find the zeros of the functions.
x1= ( -2 + sqrt(4-24) / 2 = (-2 + sqrt20*i)/2 = (-2+2sqrt5*i ) /2
==> x1= ( -2 + sqrt5*i )
==> x2= (-2 -sqrt5*i )
==> Then we conclude that the function has no real zeros. The inequality is true for all real numbers .
Also, if we draw the graph of the curve x^2 + 2x + 6 , we will notice that the curve is above the x-axis for all values.
Then the function is always positive and has no zeros.
==> The solution is :
x = R ( R is a real number)
The given inequality is x^2 + 2x + 6 >=0
The derivative of x^2 + 2x + 6 is 2x + 2
2x + 2 = 0
=> x = -1
At x = -1, x^2 + 2x + 6 = 5
Also the second derivative at x = -1 is 2 which is positive.
This shows that a minimum value of the function lies at x = -1 and it is equal to 5. As 5 > 0, the x^2 + 2x + 6 is always greater than 0.
The inequality x^2 + 2x + 6 > = 0 is valid for all real values of x.
The solution of x^2+2x+6>=0 is required.
This inequality can be rewritten in the following way:
x^2 + 2x + 1 + 5 >= 0
(x + 2)^2 + 5 >= 0
(x + 2)^2 >= -5
But the square of any number is a positive number and the inequality we have arrived at holds for all values of x.
The solution of the given inequality is the set of real numbers R.
We'll have to determine the set of values for x that makes the inequality to hold.
For this reason, we'll determine the x intercepts of the quadratic. We'll cancel the given expression
x^2+2x+6 = 0
To verify if the parabola is intercepting x axis, we'll determine the discriminant delta.
delta = b^2 - 4ac
delta = 4 - 24
delta = -20
Since delta is negative, the given expression is positive for any value of x.
The set of values of x that makes the inequality to hold is R (the real set of numbers).