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  • Stuck here, help me understand: - Complex numbers and quadratic equations - JEE Main-6

The value of 'a' for which x^{2}-11x+a= 0 and x^{2}-14x+2a= 0 have common roots are 

  • Option 1)

    0,12

  • Option 2)

    0,24

  • Option 3)

    12,24

  • Option 4)

    12,-12

 

Answers (1)

best_answer

x^{2}-11x+a=0\; \; \; and \; \; x^{2}-14x+2a=0  have common root.

\\*\therefore (1\times a-1\times 2a)^{2}=\left [ (-11)(2a)-(-14)(a) \right ]\times \left [ 1\times (-14)-1(-11) \right ]\\*\Rightarrow a^{2}=(-8a)(-3)\\*\Rightarrow a^{2}-24a=0\Rightarrow a=0\; \; or\; \; a=24

 

Condition for one common root -

\left ( {a}' c-a{c}'\right )^{2}= \left ( b{c}' -{b}'c\right )\left ( a{b}' -{a}'b\right )

- wherein

ax^{2}+bx+c=0 &

a'x^{2}+b'x+c'=0

are the 2 equations

 

 


Option 1)

0,12

This is incorrect

Option 2)

0,24

This is correct

Option 3)

12,24

This is incorrect

Option 4)

12,-12

This is incorrect

Posted by

prateek

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