Solve inequality and put solution set in interval notation.
z^2+2z+6>0
z^2+2z + 1 +5>0
(z+1)^2> -5
Which has no real solutions.
take square root of each side
z+1 < isqrt5 but sqrt5 has two roots, so
-isqrt5<z+1<+isqrr5
subtract 1 from all sides, and you have it.
-1-isqrt5<z<-1+isqrt5
I am surprised your teacher gave you this, as I suspect it is a little over your experience level. If the 6 had been -6, it would be different.
Actually, (z-1)^2 >= 0, so it is > -5 for all x!
To solve the inequality z^2 + 2z + 6 > 0, you will need to find the values of z that make the expression on the left side of the inequality greater than zero.
First, let's find the roots of the quadratic equation z^2 + 2z + 6 = 0. We can use the quadratic formula, which states that for any quadratic equation az^2 + bz + c = 0, the roots are given by:
z = (-b ± √(b^2 - 4ac)) / (2a)
For our equation z^2 + 2z + 6 = 0, a = 1, b = 2, and c = 6. Plugging these values into the quadratic formula:
z = (-(2) ± √((2)^2 - 4(1)(6))) / (2(1))
z = (-2 ± √(4 - 24)) / 2
z = (-2 ± √(-20)) / 2
Since we have a negative value under the square root (√(-20)), there are no real solutions. Therefore, the quadratic equation z^2 + 2z + 6 = 0 has no x-intercepts.
Since the parabola opens upward (the coefficient of the quadratic term is positive), it means the parabola is always above the x-axis, and the inequality z^2 + 2z + 6 > 0 holds for all values of z.
In interval notation, we can express this as (-∞, ∞) since there are no values of z that make the expression z^2 + 2z + 6 less than or equal to zero.