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  • I have a doubt, kindly clarify. - Complex numbers and quadratic equations - JEE Main-4

Let f\left ( x \right )= -x^{2}+ax+\left (a^{2}+1 \right ). Then number of values of 'a' for which f\left ( x \right )= 0 has real and equal roots is \left ( a\epsilon R \right )

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    2

  • Option 4)

    3

 

Answers (1)

best_answer

For , real & equal roots, D=0

\Rightarrow a^{2}+4(a^{2}+1)=0\Rightarrow 5a^{2}+4=0

which never holds for any real value of a so option (A)

 

Quadratic Expression Graph when a< 0 & D = 0 -

Real and equal roots of

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

& D= b^{2}-4ac

- wherein

 

 


Option 1)

0

This is correct

Option 2)

1

This is incorrect

Option 3)

2

This is incorrect

Option 4)

3

This is incorrect

Posted by

Plabita

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