Math, asked by raghavarora9414, 4 months ago

Please solve this problem​

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Answered by mathdude500
3

\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

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