Question

Please solve this problem​

Answer 1

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

Given equation is

$\red{\rm :\longmapsto\: {( \sqrt{3} + \sqrt{2}) }^{x} + {( \sqrt{3} - \sqrt{2} ) }^{x} - 2 \sqrt{3} = 0}$

Now,

Consider,

$\rm :\longmapsto\:( \sqrt{3} + \sqrt{2})( \sqrt{3} - \sqrt{2})$

$\rm \:  =  \:  \: {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2}$

$\rm \:  =  \:  \: 3 - 2$

$\rm \:  =  \:  \: 1$

$\bf\implies \: \sqrt{3} - \sqrt{2} = \dfrac{1}{ \sqrt{3} + \sqrt{2} }$

So, given equation reduces to

$\red{\rm :\longmapsto\: {( \sqrt{3} + \sqrt{2}) }^{x} + \dfrac{1}{{( \sqrt{3} + \sqrt{2} ) }^{x}} - 2 \sqrt{3} = 0}$

Now, Let assume that

$\boxed{ \: \: \: \: \: \: \bf{ {( \sqrt{3} + \sqrt{2} ) }^{x} = y - - - - (1) \: \: \: \: \: \: }}$

So, above equation reduces to

$\rm :\longmapsto\:y + \dfrac{1}{y} - 2 \sqrt{3} = 0$

$\rm :\longmapsto\:\dfrac{ {y}^{2} + 1 - 2 \sqrt{3} y}{y} = 0$

$\rm :\longmapsto\: {y}^{2} - 2 \sqrt{3}y + 1 = 0$

So, its a quadratic equation in y and to solve this quadratic, we use Quadratic Formula which is given by

$\boxed{ \bf{ y = \frac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}}}$

Here,

$\red{\rm :\longmapsto\:a = 1} \: \: \: \: \: \: \: \: \: \\ \red{\rm :\longmapsto\:b = - 2 \sqrt{3}} \\ \red{\rm :\longmapsto\:c = 1 \: \: \: \: \: \: \: \: \: }$

So, on substituting the values, we get

$\rm :\longmapsto\:y = \dfrac{ - ( - 2 \sqrt{3}) \: \pm \: \sqrt{ {( - 2 \sqrt{3} )}^{2} - 4 \times 1 \times 1} }{2 \times 1}$

$\rm :\longmapsto\:y = \dfrac{2 \sqrt{3} \: \pm \: \sqrt{ 12 - 4} }{2}$

$\rm :\longmapsto\:y = \dfrac{2 \sqrt{3} \: \pm \: \sqrt{ 8} }{2}$

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

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

$\rm :\implies\:y = \sqrt{3} + \sqrt{2} \: \: \: or \: \: \: y = \sqrt{3} - \sqrt{2}$

$\rm\implies\:{( \sqrt{3} + \sqrt{2} ) }^{x} = \sqrt{3} + \sqrt{2} \: or \:{( \sqrt{3} + \sqrt{2} ) }^{x} = \sqrt{3} - \sqrt{2}$

$\rm\implies\:{( \sqrt{3} + \sqrt{2} ) }^{x} = (\sqrt{3} + \sqrt{2})^{1} \: or \:{( \sqrt{3} + \sqrt{2} ) }^{x} = {( \sqrt{3} + \sqrt{2} )}^{ - 1}$

$\bf\implies \:x = 1 \: \: \: or \: \: \: x = - 1$

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