Solve by taking the square root of both sides 3(x-1)^2-162=0 a. x = -1 - 3 sqrt 6 or -1 + 3 sqrt 6
b. x = 1 - 3 sqrt 6 or -1 + 3 sqrt 6
c. x = -1 - 3 sqrt 6 or 1 + 3 sqrt 6
d. x = 1 - 6 sqrt 6 or 1 + 6 sqrt 6

see explanation

Step-by-step explanation:

Given

3(x - 1)² - 162 = 0 ( add 162 to both sides )

3(x - 1)² = 162 ( divide both sides by 3 )

(x - 1)² = 54 ( take the square root of both sides )

x - 1 = ± = ± 3 ( add 1 to both sides )

x = 1 ± 3

x = 1 - 3 or x = 1 + 3

clock

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.