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# If $x$ and $y$ are the roots of the equation ${x^2} + bx + 1 = 0$ , then the value of $\dfrac{1}{{x + b}} + \dfrac{1}{{y + b}}$ isA. $\dfrac{1}{b}$B. $b$C. $\dfrac{1}{{2b}}$D. $2b$

Last updated date: 04th Mar 2024
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Hint:For solving this particular question , we consider that for a quadratic equation ${x^2} + bx + 1 = 0$ , the sum of its roots $= - b$ and the product of its roots $= 1$. Therefore, $x + y = - b$ and $xy = 1$ , substitute these values to get the result .

Complete solution step by step:
It is given that , $x$ and $y$ are the roots of the equation ${x^2} + bx + 1 = 0$ ,
For a quadratic equation ${x^2} + bx + 1 = 0$ , the sum of its roots $= - b$ and the product of its roots $= 1$
Therefore, $x + y = - b$ and $xy = 1$ ,
Now ,
$\dfrac{1}{{x + b}} + \dfrac{1}{{y + b}} = \dfrac{{y + b + x + b}}{{(x + b)(y + b)}}$
$= \dfrac{{(x + y) + 2b}}{{xy + b(x + y) + {b^2}}} \\ = \dfrac{{ - b + 2b}}{{1 + b( - b) + {b^2}}} \\ = \dfrac{b}{1} \\ = b \\$
Hence , option B is the correct option.
Note: For every quadratic equation, there can be one or more than one solution. These are called the roots of the quadratic equation. , For a quadratic equation ${x^2} + bx + 1 = 0$ , the sum of its roots $= - b$ and the product of its roots $= 1$ . A quadratic equation may be expressed as a product of two binomials.