InterviewSolution
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InterviewSolution
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`x^(4) + 1 = 0` के सभी मूलों को निकालें | |
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Answer» `x^(4) + 1 = 0` `implies x^(4) + 2x^(2) + 1 - 2x^(2) = 0` `implies (x^(2) + 1)^(2) - (sqrt(2x)^(2) = 0` `implies (x^(2) + 1 + sqrt2x) ( x^(2) + 1 - sqrt2x) = 0` अब `x^(2) + sqrt2x + 1 = 0 implies x = (-2 pm sqrt(2 - 4) )/(2) = (-sqrt pm (sqrt-2))/(2)` `therefore x = (- sqrt2 pm sqrt2 i)/(2) = (-1 pm i)/(sqrt2)` पुरा `x^(2) - sqrt2x + 1 = 0` `implies x = (sqrt2 pm sqrt(2 -4 ))/(2) = (sqrt2 pm sqrt-2)/(2) = (sqrt2 pm sqrt2i)/(2) = (1 pm i)/(sqrt2)` अतः `x^(4) + 1` के सभी मूल `(-1 pm i)/(sqrt2) , (1 pm i)/(sqrt2)` है | |
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