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  • Can someone help me with this, - Complex numbers and quadratic equations - JEE Main-3

Let f\left ( x \right )= -x^{2}+\left ( m-3 \right )x-m. Then the values which m\left ( m\epsilon R \right ) can take for f\left ( x \right )= 0 to have real and distinct roots are 

  • Option 1)

    m<2

  • Option 2)

    m<3

  • Option 3)

    m>8

  • Option 4)

    m>9

 

Answers (1)

best_answer

For real and distinct roots

\rightarrow D> 0\\*\Rightarrow (m-3)^{2}-4m> 0\\*\Rightarrow m^{2}-10m+9> 0\\*\Rightarrow m< 1\cup m> 9\; \; \; \; \; \; \; \; \; \; \; (1)

(D)  is subset above intervals so option (D).

 

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)

m<2

This is incorrect

Option 2)

m<3

This is incorrect

Option 3)

m>8

This is incorrect

Option 4)

m>9

This is correct

Posted by

Plabita

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