Question 1167313
<pre>
What would be the solution for x in the rational equation:

x^2/(x-3)= (x+2)/(2x-5)

{{{matrix(1,3, x^2/(x - 3), "=", (x + 2)/(2x - 5))}}}
{{{x^2(2x - 5) = (x - 3)(x + 2)}}} ---- Cross-multiplying
{{{2x^3 - 5x^2 = x^2 - x - 6}}}
{{{2x^3 - 6x^2 + x + 6 = 0}}} <===== As the polynomial's LARGEST exponent is 3, there will be 3 solutions!!

Using the RATIONAL ROOT THEOREM, we find that 2 is a solution/zero of the above trinomial, so x = 2, or x - 2 = 0.
Hence, x - 2 is a factor of the polynomial.

Using SYNTHETIC DIVISION, or LONG DIVISION of POLYNOMIALS, the other factor of the polynomial is the quadratic, {{{2x^2 - 2x - 3}}}, so the factors of {{{2x^3 - 6x^2 + x + 6}}} are: {{{(x - 2)(2x^2 - 2x - 3)}}}.
As {{{2x^2 - 2x - 3}}} CANNOT be factored with INTEGERS, its solutions/zeroes can be found, either by COMPLETING the SQUARE
or using the QUADRATIC EQUATION formula. Using the latter, we get: 
{{{2x^2 - 2x - 3 = 0}}}
{{{x = (- b +- sqrt(b^2 - 4ac))/(2a)}}} <==== QUADRATIC EQUATION formula, where: {{{matrix(3,1, a = 2, b = - 2, c = - 3)}}}
{{{x = (- - 2 +- sqrt((- 2)^2 - 4(2)(- 3)))/(2a)}}}

{{{x = (2 +- sqrt(4 + 24))/2(2)}}}      {{{x = (2 +- sqrt(28))/4}}}          {{{x = (2 +- sqrt(4 * 7))/4}}}           {{{x = (2 +- sqrt(4)sqrt(7))/4}}}

{{{x = (2 +- 2sqrt(7))/4}}}          {{{x = (2(1 +- sqrt(7)))/2(2)}}}        {{{x = (cross(2)(1 +- sqrt(7)))/2cross((2))}}}         {{{x = (1 +- sqrt(7))/2}}} 

The 3 solutions/zeroes (1 RATIONAL and 2 IRRATIONAL): {{{highlight_green(x[1] = 2)}}}  ;  {{{highlight_green(x[2] = (1 + sqrt(7))/2)}}}  ;  {{{highlight_green(x[3] = (1 - sqrt(7))/2)}}}</pre>