# How do you solve -4/(x^2 - 11x + 30) - 7/(x^2 + x - 30) = 11/ (x^2 - 36)?

Oct 21, 2015

$x = \frac{73}{22}$.

#### Explanation:

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 $a {x}^{2} + b x + c$ are to be found with the formula

${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$.

It's evident that (using real numbers) these two solutions exist if and only if ${b}^{2} - 4 a c \setminus \ge 0$, otherwise we'll have complex solutions. In particular, if ${b}^{2} - 4 a c = 0$, 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

$a {x}^{2} + b x + c = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$.

So:

• The roots of ${x}^{2} - 11 x + 30$ are $5$ and $6$, so ${x}^{2} - 11 x + 30 = \left(x - 5\right) \left(x - 6\right)$;
• The roots of ${x}^{2} + x - 30$ are $5$ and $- 6$, so ${x}^{2} + x - 30 = \left(x - 5\right) \left(x + 6\right)$.

As for ${x}^{2} - 36$, we don't need particular calculations, because it's a special case: using the identity $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$, we derive (since ${x}^{2}$ is $x$ squared and $36$ is $6$ squared) than ${x}^{2} - 36 = \left(x + 6\right) \left(x - 6\right)$.

Now we can rewrite the initial expression as

$- \frac{4}{\left(x - 5\right) \left(x - 6\right)} - \frac{7}{\left(x - 5\right) \left(x + 6\right)} = \frac{11}{\left(x + 6\right) \left(x - 6\right)}$.

The least common denominator is thus $\left(x - 5\right) \left(x + 6\right) \left(x - 6\right)$, and we use it to write the expression with a unique, common denominator:

$- \frac{4 \left(x + 6\right)}{\left(x - 5\right) \left(x + 6\right) \left(x - 6\right)} - \frac{7 \left(x - 6\right)}{\left(x - 5\right) \left(x + 6\right) \left(x - 6\right)} = \frac{11 \left(x - 5\right)}{\left(x - 5\right) \left(x + 6\right) \left(x - 6\right)}$

Bringing the left member to the right, we obtain the following equation:

$\frac{4 \left(x + 6\right) + 7 \left(x - 6\right) + 11 \left(x - 5\right)}{\left(x - 5\right) \left(x + 6\right) \left(x - 6\right)} = 0$.

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

$4 \left(x + 6\right) + 7 \left(x - 6\right) + 11 \left(x - 5\right) = 0$

Expanding:

$4 x + 24 + 7 x - 42 + 11 x - 55 = 0$

Summing:

$22 x - 73 = 0$

Solving by $x$:

$x = \frac{73}{22}$.

Since $\frac{73}{22}$ is a good value (it doesn't annihilate the denominator), we can accept it as a solution.