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Polynomial Equations and Roots - Class XII

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The number of real solution of $x-\dfrac{1}{x^2-4}=2-\dfrac{1}{x^2-4}$ is

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$2$

💡 Explanation:

Given that:

$x-\dfrac{1}{x^2-4}=2-\dfrac{1}{x^2-4}$

$x^3-4x-1=2x^2-8-1$

$x^3-2x^2-4x+8=0$

$x^2(x-2)-4(x-2)=0$

$(x^2-4)(x-2)=0$

$(x-2)(x+2)(x-2)=0$

$(x-2)(x+2)=0$

$x=-2,+2$

Hence,

There are two real solutions for the given expression.