Question

how do i solve x2 – 2x = 8 by completing the square?

Answer 1

\[(x-1)^2=8+1=9\]
\[x-1=\pm3\]
\[x-1=3\implies x=4\]
\[x-1=-3\implies x=-2\]

Answer 2

i need to show the work, can you explain how you got this?

Answer 3

Could you @satelite try to answer my question

Answer 4

Ummm

Answer 5

We generally don't complete the square to solve for a variable. We usually use it to place a quadratic equation into vertex form in order to graph or solve a word problem.
\[x ^{2}-2x=8\]
Bring everything to one side
\[x ^{2}-2x-8=0\]
Now we place brackets around the first two terms:
\[(x ^{2}-2x)-8=0\]
Now we need to create a nice perfect square in the brackets. We do this by taking half of x's coefficient and squaring it. In this case (2/2)^2 = 1 We then add and subtract this number, thereby not "really" changing the equation.
\[(x ^{2}-2x+1-1)-8=0\]
We bring the subtracted term out of the brackets:
\[(x ^{2}-2x+1)-1-8=0\]
Then we factor the bracket:
\[(x-1)^{2}-1-8=0\]
Simplify:
\[(x-1)^{2}-9=0\]
That would be vertex form (if the 0 was a y)
So if we want to solve:
\[(x-1)^{2}=9\]
Square root both sides, this actually gives us a plus and minus version:
\[x-1= +/-3\]
so we have two solutions:
\[x=-2\] and
\[x=4\]