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  • Answer please! - Complex numbers and quadratic equations - JEE Main-4

If a,b are odd integers , then roots of equation 2ax^{2}+\left ( 2a+b \right )x+b= 0,\: a\neq 0 are

  • Option 1)

    Rational

  • Option 2)

    irrational

  • Option 3)

    Both positive

  • Option 4)

    Both -ve

 

Answers (1)

best_answer

x=\frac{-2a-b\pm \sqrt{4a^{2}+b^{2}+4ab-8ab}}{4a}

x=\frac{-2a-b\pm \left ( 2a-b \right )}{4a}

\Rightarrow \: x=\frac{-b}{2a}\: or\: x=-1

\therefore Option (A)

 

Condition for Rational roots of quadratic equation -

a,b,c\in Q ( Rational no.)

D= b^{2}-4ac   is a perfect square of rational no.

- wherein

ax^{2}+bx+c=0

is a quadratic equation

 

 


Option 1)

Rational

This is correct

Option 2)

irrational

This is incorrect

Option 3)

Both positive

This is incorrect

Option 4)

Both -ve

This is incorrect

Posted by

divya.saini

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