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  • Tell me? - Complex numbers and quadratic equations - JEE Main-4

If ax^{2}+bx+1=0  has real and equal roots \left ( Where \; a,b\epsilon R\; and\; a\neq 0 \right ) . Then ax^{2}+\left ( \sqrt{2}b \right )x+2= 0  will have

  • Option 1)

    Real & Distinct roots

  • Option 2)

    Real & equal roots

  • Option 3)

    Imaginary roots

  • Option 4)

    Data insufficient to decide

 

Answers (1)

best_answer

\because ax^{2} + bx + 1 = 0 has real and equal roots

So, b^{2} -4a =0 \Rightarrow b^{2} = 4a \;\;\;-(1)

\Delta of ax^{2} +(\sqrt{2}b)x + 2 = 0 will be 

\Delta = (\sqrt{2}b)^{2} - 4(a)(2) = 2b^{2} - 8(a) = 2(b^{2}-4a) = 0

\therefore ax^{2} +(\sqrt{2}b)x + 2 = 0 also will have real and equal roots

 

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 & Distinct roots

This is incorrect.

Option 2)

Real & equal roots

This is correct.

Option 3)

Imaginary roots

This is incorrect.

Option 4)

Data insufficient to decide

This is incorrect.

Posted by

prateek

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