math: polynomials
posted by anonymous on .
Please help me with these three problems!
1. solve for x:
3x + 5x^(1/2)  28 = 0
2. solve for x:
square root(2x5)  square root(x2) = 2
3. Find all values so that polynomial:
ax^2 + 5x + 2 has two distinct real roots.

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 = x2
2x  5 = 2y^21
sqrt (2y^2 1) = 2 + y
2y^2 1 = y^2 +4y +4
y^2 4y 5 = 0
(y5)(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