First of all, we need to deal with those denominators, factoring them by finding their roots. The roots of a quadratic equation of the form #ax^2+bx+c# are to be found with the formula
#x_{1,2} = \frac{-b\pm\sqrt(b^2-4ac)}{2a}#.
It's evident that (using real numbers) these two solutions exist if and only if #b^2-4ac\ge 0#, otherwise we'll have complex solutions. In particular, if #b^2-4ac=0#, then the two solution will collapse into the same number. Once we know the two solutions #x_1# and #x_2#, we can write the quadratic expression as
#ax^2+bx+c = (x-x_1)(x-x_2)#.
So:
- The roots of #x^2-11x+30# are #5# and #6#, so #x^2-11x+30=(x-5)(x-6)#;
- The roots of #x^2+x-30# are #5# and #-6#, so #x^2+x-30=(x-5)(x+6)#.
As for #x^2-36#, we don't need particular calculations, because it's a special case: using the identity #(a+b)(a-b)=a^2-b^2#, we derive (since #x^2# is #x# squared and #36# is #6# squared) than #x^2-36=(x+6)(x-6)#.
Now we can rewrite the initial expression as
#- 4/( (x-5)(x-6) ) - 7/( (x-5)(x+6) ) = 11 / ((x+6)(x-6))#.
The least common denominator is thus #(x-5)(x+6)(x-6)#, and we use it to write the expression with a unique, common denominator:
#- (4(x+6))/( (x-5)(x+6)(x-6) ) - (7(x-6))/( (x-5)(x+6)(x-6) ) = (11(x-5)) / ((x-5)(x+6)(x-6))#
Bringing the left member to the right, we obtain the following equation:
#(4(x+6)+7(x-6)+11(x-5))/((x-5)(x+6)(x-6))=0#.
A fraction equals zero if and only if the numerator is zero (of course the denominator cannot be zero, which we will check at the end). So, the equation reduces to
#4(x+6)+7(x-6)+11(x-5)=0#
Expanding:
#4x+24+7x-42+11x-55=0#
Summing:
#22x-73=0#
Solving by #x#:
#x=73/22#.
Since #73/22# is a good value (it doesn't annihilate the denominator), we can accept it as a solution.
