Explanation:
that means to find the value of k for which
y = x² + kx + 11
has no real number solutions in the case of y = 2.
2 = x² + kx + 11
0 = x² + kx + 9
we remember the general solution to a quadratic equation
0 = ax² + bx + c
is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = k
c = 9
x = (-k ± sqrt(k² - 4×1×9))/(2×1) =
= (-k ± sqrt(k² - 36))/2
this has only no real number solutions, if
sqrt(k² - 36)
had no real number solutions.
in other words
k² - 36 < 0
as the square root of a negative number is always a complex (or imaginary) number, and never a real value.
so,
k² < 36
|k| < 6
remember that the square root of a number has 2 solutions : one positive number and one negative number (as also the negative number multiplied by itself has the same result as the positive number multiplied by itself).
and so that means
-6 < k < 6
as interval notation that is
(-6, 6)
please consider the round brackets (parentheses), as we have to exclude the end points of the interval due to the fact that we have the "<" and not the "<=" inequalities.