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  • Can someone explain - Complex numbers and quadratic equations - JEE Main-4

If ax^{2}+2bx+b= 0 has real and distinct roots \left ( Where\; a,b\epsilon R\, ,\; a\neq 0 \right )  then ax^{2}+2\left ( a+b \right )x+3b= 0  will have roots

  • Option 1)

    Real and distinct

  • Option 2)

    Real and Equal

  • Option 3)

    Imaginary

  • Option 4)

    Depends on a & b

 

Answers (1)

best_answer

ax^{2}+2bx+b=0 has real & distinct roots so \left ( 2b \right )^{2}-4ab> 0\Rightarrow b^{2}-ab> 0

Now, for ax^{2}+2\left ( a+b \right )x+3b=0

its  D=4\left ( a+b \right )^{2}-4\left ( a \right )\left ( 3b \right )

=4\left ( a^{2}+b^{2}+2ab-3ab \right )

=4a^{2}+4\left ( b^{2}-ab \right )

=\left ( +ve \right )+\left ( +ve \right )

=+ve

\therefore D> 0

So Real and distinct roots

\therefore Option (A)

 

Condition for Real and equal roots of Quadratic Equation -

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

- wherein

ax^{2}+bx+c= 0

is the quadratic equation

 

 


Option 1)

Real and distinct

This is correct

Option 2)

Real and Equal

This is incorrect

Option 3)

Imaginary

This is incorrect

Option 4)

Depends on a & b

This is incorrect

Posted by

Aadil

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