Question

If the equation ${x^2} + bx - 1 = 0$ and ${x^2} + x + b = 0$ have a common root different from $ - 1$, then $\left| b \right|$ is equal to:
$
  A)\,\sqrt 2 \\
  B)\,3 \\
  C)\,2 \\
  D)\,\sqrt 3 \\
 $

Answer 1

Hint:we are given two quadratic equations. Now simplify those equations in order to get a common root. After getting the common root, substitute in one of the equations to get the value of $b$.

Complete step-by-step answer:
The given equations are
${x^2} + bx - 1 = 0$ $ \to (1)$
${x^2} + x + b = 0$ $ \to (2)$
Now let’s subtract equation $(2)$ from $(1)$
$
   \Rightarrow bx - x - 1 - b = 0 \\
   \Rightarrow x(b - 1) = b + 1 \\
   \Rightarrow x = \dfrac{{b + 1}}{{b - 1}} \\
 $
Therefore the common root of the equations is $\dfrac{{b + 1}}{{b - 1}}$
Hence it will depend on the value of $b$. So substituting $ \Rightarrow x = \dfrac{{b + 1}}{{b - 1}}$in (2)
\[{\left( {\dfrac{{b + 1}}{{b - 1}}} \right)^2} + \left( {\dfrac{{b + 1}}{{b - 1}}} \right) + b = 0\]
Multiplying the equation by \[{(b - 1)^2}\]
$ \Rightarrow {(b + 1)^2} + (b + 1)(b - 1) + b{(b - 1)^2} = 0$
Using properties ${(a + b)^2} = {a^2} + {b^2} + 2ab\,\,\& \,\,(a + b)(a - b) = {a^2} - {b^2}\,\,\& \,\,{(a - b)^2} = {a^2} + {b^2} - 2ab$
$
   \Rightarrow {b^2} + 1 + 2b + {b^2} - 1 + b({b^2} + 1 - 2b) = 0 \\
   \Rightarrow 2{b^2} + 3b + {b^3} - 2{b^2} = 0 \\
   \Rightarrow 3b + {b^3} = 0 \\
   \Rightarrow b({b^2} + 3) = 0 \\
   \Rightarrow b = 0\,\,\,\,or\,\,\,({b^2} + 3) = 0 \\
 $
Hence either $ \Rightarrow b = 0\,\,\,\,or\,\,\,({b^2} + 3) = 0$
$
   \Rightarrow b = 0\,\,\,\,,\,\,\,\,{b^2} = - 3 \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b = \pm i\sqrt 3 \\
 $
For $b = 0,x = - 1$
As stated in the question that $x = - 1$ cannot be the common root of the equations. Hence, $b = 0$ is rejected.
Therefore $b = \pm i\sqrt 3 $
We know that $\left |a+ib\right| = \sqrt {a^2+b^2}$
So applying above property we get,
$ \Rightarrow \left| b \right| = \sqrt 3 $ for the equations to have a common root.

So, the correct answer is “Option D”.

Note:Another method could be to assume the common root as $x = a$. Then, substitute it in the given equations. After that we will obtain two equations in variables $a\,\&\, b$. After solving we will get the values of $a\,\&\, b$. Do remember to eliminate the value of $b$ for which $x = - 1$ as it is a very common mistake.