Answer

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Hint- Solve the equation step by step by taking common and further factorizing the terms and taking common.

Given that:

$\dfrac{{x + \sqrt {{x^2} - 1} }}{{x - \sqrt {{x^2} - 1} }} - \dfrac{{x - \sqrt {{x^2} - 1} }}{{x + \sqrt {{x^2} - 1} }} = 8x\sqrt {{x^2} - 3x + 2} $

By taking LCM on the LHS and proceeding further

$ \Rightarrow \dfrac{{{{\left( {x + \sqrt {{x^2} - 1} } \right)}^2} - {{\left( {x - \sqrt {{x^2} - 1} } \right)}^2}}}{{\left( {x - \sqrt {{x^2} - 1} } \right)\left( {x + \sqrt {{x^2} - 1} } \right)}} = 8x\sqrt {{x^2} - 3x + 2} $

Breaking the terms in the numerator using algebraic identities

$

\left[ {\because {{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2} = 4ab} \right] \\

\left[ {\because \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right] \\

$

Using the above identity in the numerator and denominator respectively we get:

$

\Rightarrow \dfrac{{4x\sqrt {{x^2} - 1} }}{{{x^2} - \left( {{x^2} - 1} \right)}} = 8x\sqrt {{x^2} - 3x + 2} \\

\Rightarrow \dfrac{{4x\sqrt {{x^2} - 1} }}{1} = 8x\sqrt {{x^2} - 3x + 2} \\

\Rightarrow x\sqrt {{x^2} - 1} = 2x\sqrt {{x^2} - 3x + 2} \\

$

Bringing all of the terms of the equation in order to solve further:

\[

\Rightarrow x\sqrt {{x^2} - 1} - 2x\sqrt {{x^2} - 3x + 2} = 0 \\

\Rightarrow x\left[ {\sqrt {{x^2} - 1} - 2\sqrt {{x^2} - 3x + 2} } \right] = 0 \\

\]

Further we have two results for now that is:

$ \Rightarrow x = 0\& \left[ {\sqrt {{x^2} - 1} - 2\sqrt {{x^2} - 3x + 2} } \right] = 0$

Now we have one solution and still further solving second part:

$

\Rightarrow \sqrt {{x^2} - 1} - 2\sqrt {{x^2} - 3x + 2} = 0 \\

\Rightarrow \sqrt {{x^2} - 1} = 2\sqrt {{x^2} - 3x + 2} \\

$

Further squaring both the sides of equation, we get:

\[

\Rightarrow {x^2} - 1 = 4\left( {{x^2} - 3x + 2} \right) \\

\Rightarrow {x^2} - 1 = 4{x^2} - 12x + 8 \\

\Rightarrow 3{x^2} - 12x + 9 = 0 \\

\]

Further splitting the middle term in order to solve the equation we have:

\[

\Rightarrow 3{x^2} - 12x + 9 = 0 \\

\Rightarrow 3\left( {{x^2} - 4x + 3} \right) = 0 \\

\Rightarrow {x^2} - 4x + 3 = 0 \\

\Rightarrow {x^2} - 3x - x + 3 = 0 \\

\Rightarrow x\left( {x - 3} \right) + 1\left( {x - 3} \right) = 0 \\

\Rightarrow \left( {x + 1} \right)\left( {x - 3} \right) = 0 \\

\Rightarrow x = - 1\& x = 3 \\

\]

Hence, we have three different solutions

$x = 0, - 1,3$

Note- In order to solve such questions containing different complex and large terms, the best way to start is to manipulate the equations in order to satisfy some algebraic identities as then the questions become a bit less lengthy and easier to solve. The algebraic identities used here are mentioned along with the solution and must be remembered.

Given that:

$\dfrac{{x + \sqrt {{x^2} - 1} }}{{x - \sqrt {{x^2} - 1} }} - \dfrac{{x - \sqrt {{x^2} - 1} }}{{x + \sqrt {{x^2} - 1} }} = 8x\sqrt {{x^2} - 3x + 2} $

By taking LCM on the LHS and proceeding further

$ \Rightarrow \dfrac{{{{\left( {x + \sqrt {{x^2} - 1} } \right)}^2} - {{\left( {x - \sqrt {{x^2} - 1} } \right)}^2}}}{{\left( {x - \sqrt {{x^2} - 1} } \right)\left( {x + \sqrt {{x^2} - 1} } \right)}} = 8x\sqrt {{x^2} - 3x + 2} $

Breaking the terms in the numerator using algebraic identities

$

\left[ {\because {{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2} = 4ab} \right] \\

\left[ {\because \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right] \\

$

Using the above identity in the numerator and denominator respectively we get:

$

\Rightarrow \dfrac{{4x\sqrt {{x^2} - 1} }}{{{x^2} - \left( {{x^2} - 1} \right)}} = 8x\sqrt {{x^2} - 3x + 2} \\

\Rightarrow \dfrac{{4x\sqrt {{x^2} - 1} }}{1} = 8x\sqrt {{x^2} - 3x + 2} \\

\Rightarrow x\sqrt {{x^2} - 1} = 2x\sqrt {{x^2} - 3x + 2} \\

$

Bringing all of the terms of the equation in order to solve further:

\[

\Rightarrow x\sqrt {{x^2} - 1} - 2x\sqrt {{x^2} - 3x + 2} = 0 \\

\Rightarrow x\left[ {\sqrt {{x^2} - 1} - 2\sqrt {{x^2} - 3x + 2} } \right] = 0 \\

\]

Further we have two results for now that is:

$ \Rightarrow x = 0\& \left[ {\sqrt {{x^2} - 1} - 2\sqrt {{x^2} - 3x + 2} } \right] = 0$

Now we have one solution and still further solving second part:

$

\Rightarrow \sqrt {{x^2} - 1} - 2\sqrt {{x^2} - 3x + 2} = 0 \\

\Rightarrow \sqrt {{x^2} - 1} = 2\sqrt {{x^2} - 3x + 2} \\

$

Further squaring both the sides of equation, we get:

\[

\Rightarrow {x^2} - 1 = 4\left( {{x^2} - 3x + 2} \right) \\

\Rightarrow {x^2} - 1 = 4{x^2} - 12x + 8 \\

\Rightarrow 3{x^2} - 12x + 9 = 0 \\

\]

Further splitting the middle term in order to solve the equation we have:

\[

\Rightarrow 3{x^2} - 12x + 9 = 0 \\

\Rightarrow 3\left( {{x^2} - 4x + 3} \right) = 0 \\

\Rightarrow {x^2} - 4x + 3 = 0 \\

\Rightarrow {x^2} - 3x - x + 3 = 0 \\

\Rightarrow x\left( {x - 3} \right) + 1\left( {x - 3} \right) = 0 \\

\Rightarrow \left( {x + 1} \right)\left( {x - 3} \right) = 0 \\

\Rightarrow x = - 1\& x = 3 \\

\]

Hence, we have three different solutions

$x = 0, - 1,3$

Note- In order to solve such questions containing different complex and large terms, the best way to start is to manipulate the equations in order to satisfy some algebraic identities as then the questions become a bit less lengthy and easier to solve. The algebraic identities used here are mentioned along with the solution and must be remembered.

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