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The equation having real coefficient with one of the roots as 5+4i  is

  • Option 1)

    x^{2}-10x+41= 0

  • Option 2)

    x^{2}+10x-41= 0

  • Option 3)

    x^{2}-10x-41= 0

  • Option 4)

    x^{2}+10x+41= 0

 

Answers (1)

\because Coefficients are real so other root = 5-4i

\therefore Sum of roots = 10 & product of roots = 41

\therefore equation is x^{2}-10x+41=0

\therefore Option (A)

 

Behaviour of Imaginary roots -

For any polynomial equation with real coefficients, Imaginary roots always occur in conjugate pair.

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Option 1)

x^{2}-10x+41= 0

This is correct

Option 2)

x^{2}+10x-41= 0

This is incorrect

Option 3)

x^{2}-10x-41= 0

This is incorrect

Option 4)

x^{2}+10x+41= 0

This is incorrect

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