A value for b for which the equations
\[{x^2} + bx - 1 = 0\]
\[{x^2} + x + b = 0\],
Have one root in common is-
A.\[ - \sqrt 2 \]
B. \[i\sqrt 3 \]
C. \[i\sqrt 5 \]
D. \[\sqrt 2 \]

Hint: We assume the common root of both the equations as a variable and use that value to form equations with roots substituted in them.

Complete step by step answer:
We are given two quadratic equations:
\[{x^2} + bx - 1 = 0\] and \[{x^2} + x + b = 0\] … (1)
Let both equations have a common root ‘y’.
Since ‘y’ is the root of the equations, then it must satisfy the quadratic equations.
\[ \Rightarrow {y^2} + by - 1 = 0\] and \[{y^2} + y + b = 0\] … (2)
Now we solve for the value of common root
\[\dfrac{{{y^2}}}{{{b^2} + 1}} = \dfrac{{ - y}}{{b + 1}} = \dfrac{1}{{1 - b}}\]
Equate second and third fraction
\[ \Rightarrow \dfrac{{ - y}}{{b + 1}} = \dfrac{1}{{1 - b}}\]
Cross multiply the values from RHS and LHS
\[ \Rightarrow y = \dfrac{{ - b - 1}}{{1 - b}}\] … (3)
Now equate first and third fraction
\[ \Rightarrow \dfrac{{{y^2}}}{{{b^2} + 1}} = \dfrac{1}{{1 - b}}\]
Cross multiply the values from RHS and LHS
\[ \Rightarrow {y^2} = \dfrac{{{b^2} + 1}}{{1 - b}}\] … (4)
Since we know \[{(y)^2} = {y^2}\]
Substitute values of y and \[{y^2}\]from equations (3) and (4)
\[ \Rightarrow {\left( {\dfrac{{ - b - 1}}{{1 - b}}} \right)^2} = \dfrac{{{b^2} + 1}}{{1 - b}}\]
Cancel same terms from denominator on both sides of the equation
\[ \Rightarrow \dfrac{{{{\left( { - b - 1} \right)}^2}}}{{1 - b}} = {b^2} + 1\]
Use identity \[{(x + y)^2} = {x^2} + {y^2} + 2xy\] to solve LHS
\[ \Rightarrow \dfrac{{1 + {b^2} + 2b}}{{1 - b}} = {b^2} + 1\]
Cross multiply both sides of the equation
\[ \Rightarrow 1 + {b^2} + 2b = {b^2} + 1 - {b^3} - b\]
Cancel same terms from both sides of the equation
\[ \Rightarrow 2b = - {b^3} - b\]
Bring all terms to LHS of the equation
\[ \Rightarrow 2b + {b^3} + b = 0\]
Add terms with same powers
\[ \Rightarrow {b^3} + 3b = 0\]
Take b common from all terms in LHS
\[ \Rightarrow b({b^2} + 3) = 0\]
Equate both factors to 0
\[ \Rightarrow {b^2} + 3 = 0\] and \[b = 0\]
Shift constant to RHS
\[ \Rightarrow {b^2} = - 3\] and \[b = 0\]
Take square root on both sides
\[ \Rightarrow b = \sqrt { - 3} \] and \[b = 0\]
Since \[i = \sqrt { - 1} \]
\[ \Rightarrow b = i\sqrt 3 \] and \[b = 0\]
Since the only matching option is \[b = i\sqrt 3 \], the common root is \[b = i\sqrt 3 \].

\[\therefore \]Option B is correct.

Note: Many students make mistake of solving for the value of y from two formed equations by calculating root through determinant method, keep in mind this is a complex method and here we will get very confusing values for two equations and we will have check 4 pairs of values by equating each one separately which will take long.