Can anyone help with rational equation problems?

Question

Problem 1:

(x) over (x + 5) - (3X) over (x +2) is equal to (4) over (x^2 + 7x + 10)

** I know that I need to use the quadratic formula at the end there -- that's where I'm getting stuck.

Problem two:

(4) over (x - 1) is equal to (X) over (x + 2) - (6) over (x^2 + x - 2)

Answer 1

Problem 1
x / (x + 5) - (3x) / (x + 2) = 4 / (x + 5)(x + 2)
x (x + 2) - (3x)(x + 5) = 4
x² + 2x - 3x² - 15x - 4 = 0
- 2x² - 13x - 4 = 0
2x² + 13x + 4 = 0
x = [-13 ± √(169- 32)] / 4
x = [ -13 ± √(137) ] / 4 is exact answer.

Problem 2
4 / (x - 1) = x / (x + 2) - 6 / (x + 2)(x - 1)
4(x + 2) = x (x - 1) - 6
4x + 8 = x² - x - 6
x² - 5x - 14 = 0
(x - 7) (x + 2) = 0
x = 7 , x = - 2

Answer 2

Problem 2: you remember you must check both solutions since by multiplying both sides by the lcd, you may introduce an extraneous solution (in this case division by zero):
The least common denominator is (x+2)(x-1).
Multiplying both sides by the LCD, you get 4(x+2) = x(x-1) - 6. Simplifying, 4x + 8 = x^2 - x - 6 or 0 = x^2 - 5x - 14.
Factoring 0 = (x-7)(x+2) and solving for x you arrive at 2 solutions: x = 7 and x = -2. Checking both you find that when x = -2 you have division by zero or undefined so your only solution is x = 7.

Problem 1: Have you dealt in negative numbers under the radical? By finding the LCD of (x + 5)(x + 2) and multiplying this on both sides, you should wind up with an equation like 2x^2 + 13x + 4. Using the quadratic equation of minus b plus/minus the sq root of b^2 - 4ac divided by 2a, and substituting a = 2, b = 13, c = 4, you arrive at a solution of -13 plus/minus 2i times sq rt of 7 divided by 4.