1) Solve for x and check

Question

Radical (x^2-10x)=3i

2) Simplify

(5+ Radical 2)/(5-Radical 2)

3)describe the nature of the roots

x^2-x-6=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

Answer 1

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

Answer 2

1) square both sides
x^2 - 10x = 9i^2
x^2 - 10x = -9
x^2 - 10x +9 = 0
(x-1)(x-9) = 0
x = 1 or x=9

check (since we squared)
if x=1
LS = √(1-10) = √-9 = 3i = RS

if x=9
LS = √(81-90) = √-9 = 3i = RS

so x = 1 or x=9

Answer 3

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