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August 1, 2014

August 1, 2014

Posted by **anonymous** on Saturday, September 22, 2007 at 3:07am.

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.

- math: polynomials -
**drwls**, Saturday, September 22, 2007 at 9:23am1. 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

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