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Question: If the equation \({x^2} + bx - 1 = 0\) and \({x^2} + x + b = 0\) have a common root different from \...

If the equation x2+bx−1=0{x^2} + bx - 1 = 0 and x2+x+b=0{x^2} + x + b = 0 have a common root different from −1 - 1, then ∣b∣\left| b \right| is equal to:
A) 2 B) 3 C) 2 D) 3  A)\,\sqrt 2 \\\ B)\,3 \\\ C)\,2 \\\ D)\,\sqrt 3 \\\

Explanation

Solution

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 bb.

Complete step-by-step answer:
The given equations are
x2+bx−1=0{x^2} + bx - 1 = 0 →(1) \to (1)
x2+x+b=0{x^2} + x + b = 0 →(2) \to (2)
Now let’s subtract equation (2)(2) from (1)(1)
⇒bx−x−1−b=0 ⇒x(b−1)=b+1 ⇒x=b+1b−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 b+1b−1\dfrac{{b + 1}}{{b - 1}}
Hence it will depend on the value of bb. So substituting ⇒x=b+1b−1 \Rightarrow x = \dfrac{{b + 1}}{{b - 1}}in (2)
(b+1b−1)2+(b+1b−1)+b=0{\left( {\dfrac{{b + 1}}{{b - 1}}} \right)^2} + \left( {\dfrac{{b + 1}}{{b - 1}}} \right) + b = 0
Multiplying the equation by (b−1)2{(b - 1)^2}
⇒(b+1)2+(b+1)(b−1)+b(b−1)2=0\Rightarrow {(b + 1)^2} + (b + 1)(b - 1) + b{(b - 1)^2} = 0
Using properties (a+b)2=a2+b2+2ab  &  (a+b)(a−b)=a2−b2  &  (a−b)2=a2+b2−2ab{(a + b)^2} = {a^2} + {b^2} + 2ab\,\,\& \,\,(a + b)(a - b) = {a^2} - {b^2}\,\,\& \,\,{(a - b)^2} = {a^2} + {b^2} - 2ab
⇒b2+1+2b+b2−1+b(b2+1−2b)=0 ⇒2b2+3b+b3−2b2=0 ⇒3b+b3=0 ⇒b(b2+3)=0 ⇒b=0    or   (b2+3)=0  \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 ⇒b=0    or   (b2+3)=0 \Rightarrow b = 0\,\,\,\,or\,\,\,({b^2} + 3) = 0
⇒b=0    ,    b2=−3                           b=±i3  \Rightarrow b = 0\,\,\,\,,\,\,\,\,{b^2} = - 3 \\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b = \pm i\sqrt 3 \\\
For b=0,x=−1b = 0,x = - 1
As stated in the question that x=−1x = - 1 cannot be the common root of the equations. Hence, b=0b = 0 is rejected.
Therefore b=±i3b = \pm i\sqrt 3
We know that ∣a+ib∣=a2+b2\left |a+ib\right| = \sqrt {a^2+b^2}
So applying above property we get,
⇒∣b∣=3\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=ax = a. Then, substitute it in the given equations. After that we will obtain two equations in variables a & ba\,\&\, b. After solving we will get the values of a & ba\,\&\, b. Do remember to eliminate the value of bb for which x=−1x = - 1 as it is a very common mistake.