Find the sum of solutions of equation square root(x^2-9)=4? Solution Before finding the solutions, we need to impose constraints of existence of the square root. For the square root to exist, the radicand must be positive or equal to zero. x^2 - 9 [>=] 0 x^2 [>=] 9 The values of x that make the square root to exist belong to the intervals (- [oo] ; -3] U [3 ; + [oo] ). Now, we\'ll square raise both sides to remove the square root: x^2 - 9 = 16 We\'ll add 9 both sides: x^2 = 16 + 9 x^2 = 25 x1 = +5 and x2 = -5 Since both values of x belong to the interval of admissible values, we\'ll validate them as soutions of the given equation: x1 = -5 and x2 = 5. The requested sum of solutions of equation is x1 + x2 = -5 + 5 = 0..

1. Find the value of n if n/3 - 1/3n = 1/n
Solution
Let the equation be n/3 - 1/3n = 1/n
To solve, we will try and get rid of the denominator by multiplying by the factor.
Multiply by 3n.
==> 3n(n/3) - 3n(1/3n) = 3n(1/n)
==> n^2 - 1 = 3.
Now we will add 1 to both sides.
==> n^2 = 4.
Now we will take the root of both sides.
==> n = +-2
Then, there are two possible values for n.
==> n = { -2, 2}