Question

If -2+i√5 and 2-i√5 are two roots of a fourth degree equations form the equations

Answer 1

SOLUTION

TO DETERMINE

The four degree equation whose two roots are

-2 + i√5 and 2 - i√5

CONCEPT TO BE IMPLEMENTED

If a + ib is a roots of a quadratic equation then a - ib is another root of the same quadratic equation

EVALUATION

Here the given roots of a four degree equation are -2 + i√5 and 2 - i√5

So the all four roots are

-2 + i√5 , - 2 - i√5 , 2 - i√5 and 2 + i√5

So the required fourth degree equation is

$\sf{(x + 2 - i \sqrt{5} )(x + 2 + i \sqrt{5} )(x - 2 + i \sqrt{5} )(x - 2 - i \sqrt{5} ) = 0}$

$\sf{ \implies \bigg[{(x + 2)}^{2} + {( \sqrt{5} )}^{2} \bigg] \bigg[{(x - 2)}^{2} + {( \sqrt{5} )}^{2} \bigg] = 0}$

$\sf{ \implies ( {x}^{2} + 4x + 4 + 5)( {x}^{2} - 4x + 4 + 5) = 0 }$

$\sf{ \implies ( {x}^{2} + 4x + 9)( {x}^{2} - 4x + 9) = 0 }$

$\sf{ \implies ( {x}^{2} + 9 + 4x )( {x}^{2} + 9 - 4x ) = 0 }$

$\sf{ \implies {( {x}^{2} + 9)}^{2} - {(4x)}^{2} }$

$\sf{ \implies {x}^{4} + 18 {x}^{2} + 81 - 16 {x}^{2} = 0}$

$\sf{ \implies {x}^{4} + 2 {x}^{2} + 81 = 0}$

FINAL ANSWER

The required fourth degree equation is

$\sf{ {x}^{4} + 2 {x}^{2} + 81 = 0}$

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