# How do you solve the following? 3/(x - 3) - 4/(x - 2) = -4 √4x + 3 < 6.

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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

Posted on

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

Multiply by ( x-3)(x-2):

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

==> 3x - 6 - 4x + 12 = -4x^2 +20x - 24

==> 4x^2 - 20x - 4x + 24 - 6 + 12 = 0

==> 4x^2 - 24x + 30 = 0

Divide by 2:

==> 2x^2 - 12x + 15 = 0

==> x1= ( 12 + sqrt(144- 120) / 4

= (12 + 2sqrt6) / 4 = 3 + sqrt6 / 2

==> x2= 3 - sqrt6/ 2

√4x + 3 < 6

==> sqrt(4x) + 3 < 6

==> 2sqrtx < 3

==> sqrtx < 3/2

==> x < 9/4

Posted on

To solve (1) 3/(x - 3) - 4/(x - 2) = -4  (2)√4x + 3 < 6.

1)  3/(x-3)-4/(x-2) = -4.

We multiply both sides by LCM of senominators (x-3)(x-2).

3(x-2) -4(x-3) = -4(x-3)(x-2) =-4(x^2-5x+6)  = -4x^2+20x-24

3x-6-4x+12 = -4x^2+20x-24.

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

Subtract -4x^2+20x-24 from both sides:

4x^2-20x+24 -x+6 = 0

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

x1 = {21+sqrt(21^2-4*4*30)}/2 = {21+sqrt(-39)}/2.

x2 = {21-sqrt(-39)}/2

There are no real solution. Both  roots are x1 = {21+sqrt(-39)}/2 and x2 = {21-sqrt(-39)}/2 are imaginary.

2)

sqrt4x + 3< 6.

We subtract 3 from both sides:

sqrt4x < 6-3 = 3.

4x < 3^2 = 9.

We square both sides:

x < 9/4.

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