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  • Need explanation for: - Complex numbers and quadratic equations - JEE Main-6

If \lambda \epsilon R is such that the sum of the cubes of the roots of the equation, x^{2}+\left ( 2-\lambda \right )x +\left ( 10-\lambda \right )=0  is minimum, then the magnitude of the difference of the roots of this equation is :

  • Option 1)

    4\sqrt{2}

     

     

     

  • Option 2)

    2\sqrt{5}

  • Option 3)

    2\sqrt{7}

  • Option 4)

    20

 

Answers (1)

As we learned

\alpha _{1}\beta =\frac{\lambda -2\pm \sqrt{4-4\lambda +\lambda ^{2}-40+4\lambda }}{2}

=\frac{\left ( \lambda -2 \right )\pm \sqrt{\lambda ^{2}-36}}{2}

Difference of roots = \left | \sqrt{\lambda ^{2}-36} \right |

Also \lambda ^{3}+\beta ^{3}=\frac{\left ( \lambda -2 \right )^{3}}{4}+\frac{3\left ( \lambda -2 \right )\left ( \lambda ^{2}-36 \right )}{4}=\left ( \lambda -2 \right )\left ( \lambda ^{2} -\lambda -26\right )

Minimum value at \lambda =4

Thus difference of roots = \left | \sqrt{4^{2}-36} \right |

=\left | \sqrt{20}\: i\right |=\left | 2\sqrt{5} \: i\right |=2\sqrt{5}


Option 1)

4\sqrt{2}

 

 

 

Option 2)

2\sqrt{5}

Option 3)

2\sqrt{7}

Option 4)

20

Posted by

Vakul

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