2x-2 must be zero or greater, since we cannot have a negative quantity under the radical sign (unless we allow for imaginary roots).
Solving 2x-2≥0, we get x-1≥0, or x≥1. x must be equal to or greater than 1.
16 - 16, so the answer to the 2nd problem is the fourth one: x=4.
So even postive integers are by defention in form 2k where k is a natural number so
let the sum of even integers to n=S
S=2(1+2+3+4+5+6+7+8+......+k-1+k
divide bith sides of equation 1 by 2
0.5S=1+2+3+4+5+...........+k-1+k
S=2(k+(k-1)+..............................+2+1)
divide both sides of equation 2 by 2
0.5S=k+k-1+..............................+2+1)
by adding both we will get
___________________________
S=(k+1)(k)
so the sum will be equal to
S=

so let us test the equation
for the first 3 even number there sums will be
2+4+6=12
by our equation 3^2+3=12
gave us the same answer so our equation is correct
B. 12 would be your answer