find the solution set for the following inequality x^2 + 4 > 4x?

x^2-4x+4 = (x-2)^2 > 0
So, x can be any number except 2.

x² + 4 > 4x
x² - 2x > - 4 + (- 2)²
x² - 2x > - 4 + 4
(x - 2)² > 0
x - 2 > 0
x > 2

Answer: x > 2

x^2-4x+4>0
(x-2)^2>0
For all values of x from - infinity to + infinity,the poynomial
is >0 exept x=2,then it is =0
-inf<2 positive
x=0,poyn=0
x>2 poly>0

1.(a). x < 7 (b). x < -23/5 2.(a). x=3 (b). y=20/11 (c). 5=b 3.(A). OMIT (b). 35 units 4. My Computer wont let me draw the curve but it would be 4.

x^2 + 4 > 4x
x^2 - 4x + 4 > 0
x^2 - 2x - 2x + 4 > 0
(x^2 - 2x) - (2x - 4) > 0
x(x - 2) - 2(x - 2) > 0
(x - 2)(x - 2) > 0

x - 2 > 0
x > 2

∴ x > 2

x > 2

x^2 -4 x + 4>0
(x-2)(x-2) > 0
|x|>2