Question

solve the quadratic equation1/x+1 +2/x+2 =4/x+4 by using quadratic formula​

Answer 1

Answer:

Here is the solution

Step-by-step explanation:

it's little bit difficult.

Answer 2

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

Given equation is

$\rm :\longmapsto\:\dfrac{1}{x + 1} + \dfrac{2}{x + 2} = \dfrac{4}{x + 4}$

On taking LCM on LHS, we get

$\rm :\longmapsto\:\dfrac{x + 2 + 2(x + 1)}{(x + 1)(x + 2)} = \dfrac{4}{x + 4}$

$\rm :\longmapsto\:\dfrac{x + 2 + 2x + 2}{ {x}^{2} + x + 2x + 2} = \dfrac{4}{x + 4}$

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

$\rm :\longmapsto\:4( {x}^{2} + 3x + 2) = (3x + 4)(x + 4)$

$\rm :\longmapsto\:4{x}^{2} + 12x + 8 = {3x}^{2} + 12x + 4x + 16$

$\rm :\longmapsto\:4{x}^{2} + 12x + 8 = {3x}^{2} + 16x + 16$

$\rm :\longmapsto\:4{x}^{2} + 12x + 8 - {3x}^{2} - 16x - 16 = 0$

$\rm :\longmapsto\:{x}^{2} - 4x - 8 = 0$

Now, we know Quadratic Formula,

$\red{\rm :\longmapsto\:Solution \: of \: quadratic \: equation \: {ax}^{2} + bx + c = 0 \: is \: }$

$\purple{ \boxed{ \bf{ \: \: \: x \: = \: \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac } }{2a} \: \: \: }}}$

Now, we have

$\rm :\longmapsto\:a = 1$

$\rm :\longmapsto\:b = - \: 4$

$\rm :\longmapsto\:c = - \: 8$

So, on substituting the values in Quadratic Formula, we get

$\rm :\longmapsto\:x \: = \: \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}$

$\rm :\longmapsto\:x \: = \: \dfrac{ - ( - 4) \: \pm \: \sqrt{ {( - 4)}^{2} - 4(1)( - 8) } }{2(1)}$

$\rm :\longmapsto\:x \: = \: \dfrac{ 4 \: \pm \: \sqrt{ 16 + 32} }{2}$

$\rm :\longmapsto\:x \: = \: \dfrac{ 4 \: \pm \: \sqrt{ 48} }{2}$

$\rm :\longmapsto\:x \: = \: \dfrac{ 4 \: \pm \: \sqrt{ 4 \times 4 \times 3} }{2}$

$\rm :\longmapsto\:x \: = \: \dfrac{ 4 \: \pm \:4 \sqrt{ 3} }{2}$

$\bf\implies \:x = 2 \: \pm \: 2 \sqrt{3}$

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