Math
posted by JP .
1) Solve for x and check
Radical (x^210x)=3i
2) Simplify
(5+ Radical 2)/(5Radical 2)
3)describe the nature of the roots
x^2x6=0
x^2+4x+29=0
4) Solve for x and check
Radical (x^2 +4x +44) +3=2x
I got 7 and 5/3 as answers, was not sure though

2) Use the fact that (a + b)(a  b) = a^2  b^2
3) Let's say your equation in the format ax^2 + bx + c = 0
In your two cases, a = 1.
Calculate the quantity b^2  4ac for each equation. If it is positive there are two real roots. If it is zero there is one. If it is negative, there are two complex roots 
1) square both sides
x^2  10x = 9i^2
x^2  10x = 9
x^2  10x +9 = 0
(x1)(x9) = 0
x = 1 or x=9
check (since we squared)
if x=1
LS = √(110) = √9 = 3i = RS
if x=9
LS = √(8190) = √9 = 3i = RS
so x = 1 or x=9 
4) since you clearly squared both sides to solve, each answer you obtained has to be verified in the original equation
if x=7 it works
if x= 5/3 it does not work
so x = 7 is the only solution
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