X^2-18x+81=0
Can someone please help?!?!?!?
note that 81 = 9^2 and 18 = 2*9
That means you have a perfect square:
(x-9)^2 = 0
So the roots are ...
So it would be
(X-9) (x+9)?
So then x=9
x^2-18x+81 = 0. 81 = -9*-9. (-9)+(-9) = -18 - B.
(x-9)(x-9) = 0
x-9 = 0, X = 9.
Only one solution.