PLEASE HELP! Solve by completing the square:
x^2-6x+8=0

Question

PLEASE HELP! Solve by completing the square: 
x^2-6x+8=0

anonymous · Answer

(x + a)(x+ b) = x^2 +(a+b)x +ab

anonymous · Answer

so (x-2)(x-4)

solomonzelman · Answer

$$x^2-6x+8=0$$$$x^2-6x+9=1$$$$(x-3)^{2}=1$$$$x-3=±1$$ FINISH...

anonymous · Answer

wow 
lets try this
start with 
$$x^2-6x=-8$$ then since half of 6 is 3, and 3 squared is 9, to right to 
$$(x-3)^2=-8+9=1$$ and go from there
or the method of @SolomonZelman  either way

solomonzelman · Answer

Well... what you said, @satellite73 is how my teacher taught me, but I only took what I did from that ;)

wolf1728 · Answer

x² -6x +8=0
Move non-x term to the right
x² -6x =-8
Divide the equation by the coefficient of X²
The coefficient is 1 so this doesn't have to be done
x² -6x =-8
take the coefficient of X which is -6 
divide that by 2 
which is -3
square that number 
9
add it to both sides of the equation
x² -6x +9= 1
Take square root of both sides
x-3 = 1
x = 2

wolf1728 · Answer

x = 4

anonymous · Answer

Thanks guys :) Much appreciated. Wish I could give more than one medal