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  • Can someone explain - Complex numbers and quadratic equations - JEE Main-3

If \left ( a^{2}+1 \right )x^{2}+2\left ( a+1 \right )x+1=0  has real and distinct roots then 

  • Option 1)

    a< 0

  • Option 2)

    a< 5

  • Option 3)

    a> 0

  • Option 4)

    a< -1

 

Answers (1)

best_answer

For real and distinct roots , D> 0

\therefore \left ( 2\left ( a+1 \right ) \right )^{2}-4(a^{2}+1)-4> 0

\Rightarrow 4a^{2}+8a+4-4a^{2}-4> 0

\Rightarrow a> 0

Condition for Real and distinct roots of Quadratic Equation -

D= b^{2}-4ac> 0

- wherein

ax^{2}+bx+c= 0

is the quadratic equation

 

 


Option 1)

a< 0

This is incorrect

Option 2)

a< 5

This is incorrect

Option 3)

a> 0

This is correct

Option 4)

a< -1

This is incorrect

Posted by

prateek

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