Jake is solving x^2-6x+5=0 by completing the square his steps are shown below step 1. x^2-6x=-5 step 2. x^-6x+9=-5+9 step 3. (x-6)^2=4 Step 4. sq root (x-6)^2=sq root 4 step 5. x-6=+- 2 step 6. x=6+-2 step 7. x=8 and x =4. which step is Jakes error in.

Jake's error in step 3

Step-by-step explanation:

we have

Complete the square

step 1

Group terms that contain the same variable, and move the constant to the opposite side of the equation

step 2

Complete the square. Remember to balance the equation by adding the same constants to each side

step 3

Rewrite as perfect squares

Jake's error in step 3

He placed 6 instead of 3 in the left side

step 4

take square root both sides

step 5

step 6

step 7

x=5 and x=1

answer: the common difference is -0.10

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'.