Question

please solve these questions ​

Answer 1

Answer:

$\displaystyle \sf \longrightarrow x=\frac{-1\pm\sqrt{17}}{2}$

Step-by-step explanation:

Given :

$\displaystyle \sf \dfrac{1}{2x-3} -\dfrac{1}{2x+5} =8$

Solving it further we get :

$\displaystyle \sf \longrightarrow \dfrac{(2x+5)-(2x-3)}{(2x-3)(2x+5)} =8$

$\displaystyle \sf \longrightarrow \dfrac{(2x+5-2x+3)}{(2x-3)(2x+5)} =8$

$\displaystyle \sf \longrightarrow \dfrac{(8)}{(2x-3)(2x+5)} =8$

$\displaystyle \sf \longrightarrow \dfrac{1}{(2x-3)(2x+5)} =1$

$\displaystyle \sf \longrightarrow (2x-3)(2x+5) =1$

$\displaystyle \sf \longrightarrow (4x^2+10x-6x-15) =1$

Adding - 1 both side we get :

$\displaystyle \sf \longrightarrow (4x^2+10x-6x-15-1) =1 -1$

$\displaystyle \sf \longrightarrow (4x^2+4x-16) =0$

$\displaystyle \sf \longrightarrow 4(x^2+x-4)=0$

Dividing by 4 both side we get :

$\displaystyle \sf \longrightarrow (x^2+x-4)=0$

Now roots of equation is given as :

$\displaystyle \sf \longrightarrow x=\frac{-b\pm\sqrt{b^4-4ac}}{2a}$

$\displaystyle \sf \longrightarrow x=\frac{-1\pm\sqrt{1^2-4\times(-4)\times1}}{2\times1}$

$\displaystyle \sf \longrightarrow x=\frac{-1\pm\sqrt{1+16}}{2}$

$\displaystyle \sf \longrightarrow x=\frac{-1\pm\sqrt{17}}{2}$

Hence we get required answer.