Question

Find the roots of the following equations :

$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30},x\ne -4,7$

Answer 1

Step 1:
$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$
$\Rightarrow \frac{\left[\right(x-7)-(x+4\left)\right]}{(x+4)(x-7)}=\frac{11}{30}$
$\Rightarrow -\frac{11}{(x+4)(x-7)}=\frac{11}{30}$
$\Rightarrow (x+4)(x-7)=-30$
$\Rightarrow x^2-3x-28=-30$
$\Rightarrow x^2-3x+2=0$

Step 2:
We know that, The standard equation is
$a{x}^2+bx+c=0$
And the obtained equation is
$x^2-3x+2=0$
On comparing we get, $a=1,b=-3,c=2$

Step 3:
Now, $a\times c=2=(-1)(-2)$
and $b=-3(-1-2)$

Step 4:
$\Rightarrow x^2-x-2x+2=0$

Step 5:
$\Rightarrow x^2-x-2x+2=0$
$\Rightarrow x(x-1)-2(x-1)=0$
$\Rightarrow (x-1)(x-2)=0$ (Grouping the term)

Step 6:
$\Rightarrow (x-1)(x-2)=0$
Thus, either $x-1=0$ or $x-2=0$
$\Rightarrow x=1\text{or}x=2$
$\therefore x=1\text{or}x=2$