Solve the inequality x^2+2x+6>=0.

Expert Answers
hala718 eNotes educator| Certified Educator

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)

justaguide eNotes educator| Certified Educator

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.

tonys538 | Student

The solution of x^2+2x+6>=0 is required.

This inequality can be rewritten in the following way:

x^2+2x+6>=0

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.

giorgiana1976 | Student

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).

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