Root of the equation x-1-a2-b2-0 is

The correct option is B b 1/a (x+1/a)^2 b^2 =0 ⇒ (x+1/a)^2 =b^2 ⇒ (x+1/a)= ± b ⇒ x =(± b) 1/a So, the roots of the equation (x+1/a)^2 b^2 =0 is ( b 1/a) an ...

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Standard X

Mathematics

Roots of Quadratic Equation

Root of the e...

Question

Root of the equation $(x + \frac{1}{a})^{2} - b^{2} = 0 i s$

$\frac{- b}{a}$

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$- a b$

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$- b - \frac{1}{a}$

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$b + \frac{1}{a}$

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Solution

The correct option is C
$- b - \frac{1}{a}$

$(x + \frac{1}{a})^{2} - b^{2} = 0 \Rightarrow (x + \frac{1}{a})^{2} = b^{2} \Rightarrow (x + \frac{1}{a}) = \pm b \Rightarrow x = (\pm b) - \frac{1}{a}$

So, the roots of the equation $(x + \frac{1}{a})^{2} - b^{2} = 0$ is $(- b - \frac{1}{a})$ and $(b - \frac{1}{a})$

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Root of the equation (x+1a)2−b2=0 is

The correct option is C −b−1a (x+1a)2−b2=0⇒(x+1a)2=b2⇒(x+1a)=±b⇒x=(±b)−1a So, the roots of the equation (x+1a)2−b2=0 is (−b−1a) and (b−1a)

Root of the equation $(x + \frac{1}{a})^{2} - b^{2} = 0 i s$