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  • Help me solve this - Complex numbers and quadratic equations - JEE Main-13

Let p,q\; \epsilon\; \textbf{R}. If 2-\sqrt{3} is a root of the quadratic equation,x^{2}+px+q=0, then :

  • Option 1)

     p^{2}-4q+12=0          

  • Option 2)

     q^{2}-4p-16=0

  • Option 3)

    q^{2}+4p+14=0

  • Option 4)

    p^{2}-4q-12=0

 

Answers (1)

best_answer

Given that one root is 2-\sqrt{3}

then another root will by 2+\sqrt{3}

Sum of roots = \alpha +\beta =4=-p

Product of roots= 2\beta =1=q

\therefore p^{2}-4q-12=0

Question is wrong

 


Option 1)

 p^{2}-4q+12=0          

Option 2)

 q^{2}-4p-16=0

Option 3)

q^{2}+4p+14=0

Option 4)

p^{2}-4q-12=0

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