In simplest radical form, what are the solutions to the quadratic equation 6 = x2 – 10x?

quadratic formula: x = startfraction negative b plus or minus startroot b squared minus 4 a c endroot over 2 a endfraction

x = 5plus or minus startroot 31 endroot
x = 5plus or minus startroot 19 endroot
x = 5plus or minus 2 startroot 19 endroot
x = 5plus or minus 2 startroot 31 endroot

x = 5 + √31 amd x = 5 - √31

Step-by-step explanation:

Rewrite 6 = x2 – 10x using " ^ " to denote exponentiation.  Then re-arrange the terms in descending powers of x:

x^2 – 10x - 6 = 0

Let's solve this by completing the square.  Add 25 to both sides and simplify the result:

x^2 - 10x + 25 = 6 + 25 = 31

The first three terms can be rewritten as

(x - 5)^2

and so we now have (x - 5)^2 = 31.

Taking the sqrt of both sides, we get

x - 5 = ± √31

And so the two real, unequal roots are

x = 5 + √31 amd x = 5 - √31

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'.

21

jk it's 20

step-by-step explanation: