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  • Can someone help me with this, - Complex numbers and quadratic equations - JEE Main-2

Let f\left ( x \right )= 2x^{2}+ax+2. Then values of 'a' for which f\left ( x \right )= 0 doesn't have imaginary roots are

  • Option 1)

    \left |a \right |\geq 1

  • Option 2)

    \left |a \right |\geq2

  • Option 3)

    \left |a \right |\geq4

  • Option 4)

    \left |a \right |\geq3

 

Answers (1)

best_answer

\because f(x)=0  doesn't have imaginary roots so either roots are real and equals or roots are real and distinct so D\geq 0

\\*\therefore a^{2}-16\geq 0\\*\Rightarrow a\leq -4\cup a\geq 4\\*\Rightarrow \left | a \right |\geq 4

 

Quadratic Expression Graph when a> 0 & D > 0 -

Real and distinct roots of

f\left ( x \right )= ax^{2}+bx+c

& D= b^{2}-4ac

- wherein

 

 


Option 1)

\left |a \right |\geq 1

This is incorrect

Option 2)

\left |a \right |\geq2

This is incorrect

Option 3)

\left |a \right |\geq4

This is correct

Option 4)

\left |a \right |\geq3

This is incorrect

Posted by

divya.saini

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