How do you solve the following? 3/(x - 3) - 4/(x √4x + 3

We have to solve

• 3/(x -
  3) - 4/(x - 2) = -4

• sqrt (4x) + 3 <
  6.

Now 3/(x - 3) - 4/(x - 2) =
-4

multiply all terms by
(x-3)(x-2)

=> 3(x-2) - 4(x-3) =
-4(x-3)(x-2)

=> 3x - 6 - 4x + 12 = -4 ( x^2 - 5x +
6)

=> 6 - x = -4x^2 + 20x -
24

=> 4x^2 - 21x + 30 =0

Now
find the roots of 4x^2 - 21x + 30 =0 using

[–b + sqrt (b^2 – 4ac)]/
2a and [–b - sqrt (b^2 – 4ac)]/ 2a

here b = -21, a = 4 and c =
30

sqrt (b^2 - 4ac) = sqrt
-39

Therefore the roots are 21/8 - i*(sqrt 39)/8 and
21/8 + i*(sqrt 39)/8.

sqrt (4x) + 3
< 6

=> sqrt 4x <
3

=> sqrt 4x < sqrt
9

=> 4x < 9

=> x
< 9/4

Therefore x<
9/4