Question

4/x-8=2/x-3+3/x+2 solve by quadratic formula

Answer 1

$\large\underline{\sf{Solution-}}$

The given quadratic equation is

$\rm :\longmapsto\:\dfrac{4}{x - 8} = \dfrac{2}{x - 3} + \dfrac{3}{x + 2}$

$\rm :\longmapsto\:\dfrac{4}{x - 8} = \dfrac{2(x + 2) + 3(x - 3)}{(x - 3)(x + 2)}$

$\rm :\longmapsto\:\dfrac{4}{x - 8} = \dfrac{2x + 4 + 3x - 9}{ {x}^{2} + 2x - 3x - 6}$

$\rm :\longmapsto\:\dfrac{4}{x - 8} = \dfrac{5x - 5}{ {x}^{2} - x - 6}$

$\rm :\longmapsto\:4( {x}^{2} - x - 6) = (x - 8)(5x - 5)$

$\rm :\longmapsto\:4{x}^{2} - 4x -24 = 5 {x}^{2} - 40x - 5x + 40$

$\rm :\longmapsto\:4{x}^{2} - 4x -24 = 5 {x}^{2} - 45x + 40$

$\rm :\longmapsto\:4{x}^{2} - 4x -24 - 5 {x}^{2} + 45x - 40 = 0$

$\rm :\longmapsto\: - {x}^{2} + 41x - 64 = 0$

$\rm :\longmapsto\: - ({x}^{2} - 41x + 64 )= 0$

$\rm :\longmapsto\: {x}^{2} - 41x + 64= 0$

Here,

$\red{\rm :\longmapsto\:a = 1}$

$\red{\rm :\longmapsto\:b = - 41}$

$\red{\rm :\longmapsto\:c = 64}$

So,

Let first find Discriminant, D.

$\rm :\longmapsto\:D = {b}^{2} - 4ac$

$\rm \: = \: {( - 41)}^{2} - 4 \times 1 \times 64$

$\rm \: = \:1681 - 256$

$\rm \: = \:1425$

$\bf\implies \:D = 1425 > 0$

So, Solution of quadratic equation is given by

$\rm :\longmapsto\:x = \dfrac{ - b \: \pm \: \sqrt{D} }{2a}$

$\rm \: = \: \dfrac{ - ( - 41) \: \pm \: \sqrt{1425} }{2 \times 1}$

$\rm \: = \: \dfrac{ 41 \: \pm \: \sqrt{5 \times 5 \times 57} }{2 }$

$\rm \: = \: \dfrac{ 41 \: \pm \: 5 \sqrt{ 57} }{2 }$

Hence,

$\bf\implies \:x \: = \: \dfrac{ 41 \: \pm \: 5 \sqrt{ 57} }{2 }$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac