Please help me with these three problems!

1. solve for x:
3x + 5x^(1/2) - 28 = 0

2. solve for x:
square root(2x-5) - square root(x-2) = 2

3. Find all values so that polynomial:
ax^2 + 5x + 2 has two distinct real roots.

1 answer

1. Treat y = sqrt x as the variable ans solve for y. Then factor or use the quadratic equation.
3y^2 + 5y -28 = 0
(3y -7)(y +4) = 0
y = -4 x = +/- 2i are two of the answers. You do the other two.

2. let y^2 = x-2
2x - 5 = 2y^2-1
sqrt (2y^2 -1) = 2 + y
2y^2 -1 = y^2 +4y +4
y^2 -4y -5 = 0
(y-5)(y+1) = 0
y = 5 or -1

Now solve for x

3. Require that b^2 - 4ac > 0
(b = 5 and c = 2).
Solve the inequality for a