Question

form a cubic equation whose roots are 1,i

Answer 1

SOLUTION

TO DETERMINE

The cubic equation whose roots are 1,i

CONCEPT TO BE IMPLEMENTED

If a + ib is a root of an equation then a - ib is also a root of the equation

EVALUATION

Here it is given that '' i '' is a root of the equation

So '' - i '' is also a root of the equation

So the three roots of the cubic equation are 1, i, - i

Hence the cubic equation is

$\sf{}(x - 1)(x + i)(x - i) = 0$

$\implies \sf{}(x - 1)( {x}^{2} - {i}^{2} ) = 0$

$\implies \sf{}(x - 1)( {x}^{2} + 1) = 0 \: \: ( \: \because \: {i}^{2} = - 1)$

$\implies \sf{} {x}^{3} + x - {x}^{2} - 1 = 0$

$\implies \sf{} {x}^{3} - {x}^{2} + x- 1 = 0$

FINAL ANSWER

Hence the required cubic equation is

$\boxed{ \sf{} \: \: {x}^{3} - {x}^{2} + x- 1 = 0 \: \: }$

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Answer 2

form a cubic equation whose roots are 1,i