Since the 128 lies between 121 and 144, it's square root lies between -12 and -11, given by option b.
The squares of numbers, until a square is greater than 128, is:
[tex](\pm 0)^2 = 0[/tex]
[tex](\pm 1)^2 = 1[/tex]
[tex](\pm 2)^2 = 4[/tex]
[tex](\pm 3)^2 = 9[/tex]
[tex](\pm 4)^2 = 16[/tex]
[tex](\pm 5)^2 = 25[/tex]
[tex](\pm 6)^2 = 36[/tex]
[tex](\pm 7)^2 = 49[/tex]
[tex](\pm 8)^2 = 64[/tex]
[tex](\pm 9)^2 = 81[/tex]
[tex](\pm 10)^2 = 100[/tex]
[tex](\pm 11)^2 = 121[/tex]
[tex](\pm 12)^2 = 144[/tex]
128 is between 121 and 144, thus, it's square root is between [tex]\pm 11[/tex] and [tex]\pm 12[/tex].
We are working with negative numbers, thus between -12 and -11, option b.
A similar problem is given at https://brainly.com/question/3729492
