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QuestionSolve $\frac{8}{x+2}-\frac{29}{x-5}-\frac{4}{\left(x+2\right)\left(x-5\right)}=10$
Note that $x \neq-2$ and $x \neq5$
Answer$\frac{8}{x+2}-\frac{29}{x-5}-\frac{4}{\left(x+2\right)\left(x-5\right)}=10$
Or, $8\left(x-5\right)-29\left(x+2\right)-4=10\left(x+2\right)\left(x-5\right)$
Or, $8x-40-29x-58-4=10\left(x^{2}-3x-10\right)$
Or, $-21x-102=10x^{2}-30x-100$Or, $-21x-102-10x^{2}+30x+100=0$Or, $-10x^{2}+9x-2=0$Or, $10x^{2}-9x+2=0$
The solutions of a
quadratic equation $ax^2+bx + c = 0$ are given by$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
In this
quadratic equation, $a=10$, $b=-9$, and $c=2$
Substituting these values of $a$, $b$ and $c$ in the above formula, we get
$x=\frac{9\pm \sqrt{81-80}}{20}$
or, $x = \frac{9\pm \sqrt{1}}{20}$
or, $x = \frac{9\pm 1}{20}$
or, $x = \frac{10}{20}$ and $x = \frac{8}{20}$
or, $x=\frac{1}{2}$ and $x =\frac{2}{5}$
There are two distinct solutions given by $x = \frac{1}{2}$ and $x = \frac{2}{5}$