Which equation has the solutions x = 1+
5?
x2 + 2x + 4 = 0
x2 - 2x + 4 = 0
0x²+2x–4=0
x2 - 2x -4 = 0

The last equation x2 - 2x -4 = 0

has solution  (x - 1)^2 - 5 = 0,   x  =  1 + root(5)  or  x = 1 - root(5)

Step-by-step explanation:

If a quadratic function has roots 1 and 5

f(x) = (x -1)(x- 5)

f(x) = x^2 - 6x + 5

Unless you meant.  -4 and 6  ?

g(x) = (x + 4)(x - 6)

g(x) = x^2 -2x -24

Or did you mean  x = 1 and x =4 ?...

x^2 + 2x + 4 = 0  :   complete square  x^2 + 2x + 1 + 3 = 0,   (x+1)^2 + 3 = 0

x^2 - 2x + 4 = 0 :  complete square:  (x -1)^2 + 3 = 0

0x^2 + 2x - 4 = 0,    2x - 4 = 0,  x = 2

x^2 - 2x - 4 = 0  becomes:   x^2 - 2x + 1 - 1 -4 = 0 ;  (x - 1)^2 - 5 = 0

can i see the following ?

step-by-step explanation:

30005

yes ok