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+cax2+bx+c are to be found with the formula
x_{1,2} = \frac{-b\pm\sqrt(b^2-4ac)}{2a}x1,2=−b±√b2−4ac2a.
It's evident that (using real numbers) these two solutions exist if and only if b^2-4ac\ge 0b2−4ac≥0, otherwise we'll have complex solutions. In particular, if b^2-4ac=0b2−4ac=0, then the two solution will collapse into the same number. Once we know the two solutions x_1x1 and x_2x2, we can write the quadratic expression as
ax^2+bx+c = (x-x_1)(x-x_2)ax2+bx+c=(x−x1)(x−x2).
So:
- The roots of x^2-11x+30x2−11x+30 are 55 and 66, so x^2-11x+30=(x-5)(x-6)x2−11x+30=(x−5)(x−6);
- The roots of x^2+x-30x2+x−30 are 55 and -6−6, so x^2+x-30=(x-5)(x+6)x2+x−30=(x−5)(x+6).
As for x^2-36x2−36, we don't need particular calculations, because it's a special case: using the identity (a+b)(a-b)=a^2-b^2(a+b)(a−b)=a2−b2, we derive (since x^2x2 is xx squared and 3636 is 66 squared) than x^2-36=(x+6)(x-6)x2−36=(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))−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)(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))−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))=04(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)=04(x+6)+7(x−6)+11(x−5)=0
Expanding:
4x+24+7x-42+11x-55=04x+24+7x−42+11x−55=0
Summing:
22x-73=022x−73=0
Solving by xx:
x=73/22x=7322.
Since 73/227322 is a good value (it doesn't annihilate the denominator), we can accept it as a solution.
