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

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Last updated date: 04th Mar 2024
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IVSAT 2024
Answer
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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.
Additional Information:
A quadratic could be a polynomial whose highest power is that of the square of a variable. Every quadratic gives two values of the unknown variable and these values are called roots of the equation. A quadratic has two roots which can be unequal real numbers or equal real numbers, or numbers which don't seem to be real. we'll solve a quadratic within the following way:
(i) First we'd like to specific the given equation within the general style of the quadratic then (ii) we want to factorize the left side of the equation, (iii) Now express each of the 2 factor equals to zero and solve them (iv)The two solutions are called the roots of the given equation.

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.
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