# Find all real solution to the rational equation x / (x + 4) = -3 / (x - 2) + 18 / (x - 2) (x + 4)

Asked on by lalooo

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

x/ (x+ 4)  = -3 / (x-2) +  18/ (x-2)(x+4)

First we will find the common denominator for all three terms:

the common denominator is : ( x-2)(x+4)

==> x(x-2) / (x-2)(x+4) = -3(x+4)/ (x-2)(x+4)  + 18/(x-2)(x+4)

Now we can reduce the denominator because it is common for all terms:

==> x(x-2) = -3(x+4) + 18

Now we will expand brackets:

==> x^2 - 2x = -3x - 12 + 18

Now we will combine like terms:

==> x^2 - 2x = -3x + 6

==> x^2 - 2x + 3x - 6 = 0

==> x62 + x - 6 = 0

Now we will factor the quadratic functions:

==> (x + 3) (x-2) = 0

==> x1= -3

==> x2= 2 ( we will not consider this solution because the function is not defined ( the denominator is 0 )

Then the only solution is x = -3

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the real solutions of  x / (x + 4) = -3 / (x - 2) + 18 / (x - 2) (x + 4).

Both sides of the given equation have denominators whose LCM = (x-2)(x+4). Therefore we multiply both sides by the LCM.

x(x-2) = 3(x+4) +18.

x^2-2x = 3x+12+18.

x^2-2x -3x = 30.

x^2-5x-30 = 0.

We use the formula to find the  roots of this quadratic equation:

x = {5+sqrt(5^2- 4*1*(-30)}/2 = (5+sqrt125)/2 = 5(1+sqrt5)/2.

x2 =5(1-sqrt5)/2.

Therefore 5(1+sqrt5)/2 and 5(1-sqrt5)/2 are the real roots of the given equation.

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