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  • I have a doubt, kindly clarify. - Complex numbers and quadratic equations - JEE Main-8

If the equations x2 + bx − 1 = 0 and x2 + x + b = 0 have a common root different from −1, then \left | b \right | is equal to :

 

  • Option 1)

    \sqrt{2}

  • Option 2)

    2

  • Option 3)

    3

  • Option 4)

    \sqrt{3}

 

Answers (2)

best_answer

As we have learned

Condition for one common root -

\left ( {a}' c-a{c}'\right )^{2}= \left ( b{c}' -{b}'c\right )\left ( a{b}' -{a}'b\right )

- wherein

ax^{2}+bx+c=0 &

a'x^{2}+b'x+c'=0

are the 2 equations

 

 For one common root , we have  , 

(c_1a_2-c_2a_1)^2= (a_1b_2-a_2b_1) (b_1c_2-b_2c_1)

\Rightarrow (-1 \cdot 1-b\cdot 1^2) = (1\cdot 1-1\cdot b)\cdot (b\cdot b-1(-1))

\Rightarrow (1+b)^2 = (1-b)(b^2+1)

\Rightarrow 1+b^2+2b = b^2+1-b^3-b

\Rightarrow b^3 +3b= 0

b= 0 , b^2= -3 \Rightarrow b= \sqrt3i \Rightarrow |b|= \sqrt3

 

 

 

 

 


Option 1)

\sqrt{2}

Option 2)

2

Option 3)

3

Option 4)

\sqrt{3}

Posted by

Himanshu

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